^1 


%^ 


IN  MEMORIAM 
FLORIAN  CAJORl 


J^'atc    (^^>^i^r^ 


PLANE   TRIGONOMETRY 


BY 


JAMES  M.  TAYLOR,  A.M.,  LL.D. 
Professor  of  Mathematics,  Colgate  University 


BOSTON,  U.S.A. 

GINN   &   COMPANY,  PUBLISHERS 

C^e  3ttl)cnaciim  prefiSfi 

1904 


Copyright,  1904 
By  JAMES  M.  TAYLOR 


ALL  BIGHTS  RESERVED 


V 


T-5 


PREFACE 


This  book  is  designed  to  meet  the  needs  of  beginners  who 
wish  to  master  the  fundamental  principles  of  Trigonometry. 
The  author's  aim  has  been  to  prepare  a  text-book  which  shall 
be  clear  and  practical,  yet  thoroughly  scientific. 

The  proofs  of  formulas  are  simple  but  rigorous.  The  use  of 
directed  lines  is  consistent;  the  directions  of  such  lines  in  the 
figures  are  usually  indicated  by  arrowheads,  and  these  lines 
are  always  read  from  origin  to  end.  Both  trigonometric  ratios 
and  trigonometric  lines  are  employed,  but  at  first  the  ratios 
are  used  exclusively  until  they  have  become  fixed  in  mind  and 
familiar.  It  is  proved  that  the  ratios  are  the  measures  of  the 
lines,  and  that,  therefore,  any  relation  which  holds  true  for 
the  one  holds  true  for  the  other  also. 

The  distinction  between  identities  and  equations  is  empha- 
sized in  definition,  treatment,  and  notation.  The  solution  of 
trigonometric  equations  is  scientific  and  general.  The  trigo- 
nometric ratios  are  defined  in  pairs  as  reciprocals  of  each 
other  both  to  aid  the  memory  and  to  emphasize  one  of  the 
most  important  of  their  fundamental  relations.  In  the  reduc- 
tion of  the  functions  of  any  angle  to  those  of  an  acute  angle, 
the  theorems  concerning  the  functions  of  —  yl  and  90°  -f  A  are 
made  fundamental.  The  addition  formulas  are  proved  for 
positive  or  negative  angles  of  any  quadrant  and  from  them 
are  deduced  the  other  formulas  concerning  the  functions  of 
two  or  more  angles.  When  two  or  more  figures  are  used  in 
a  proof,  the  same  phraseology  always  applies  to  each  figure. 


M«n«i>4A 


iv  • PREFACE 

In  the  first  chapter  will  be  found  a  table  of  natural  functions 
at  intervals  of  5°,  which  the  student  is  to  verify  to  two  places 
by  construction  and  measurement,  and  many  interesting  prob- 
lems have  been  introduced  in  order  that  the  subject  may  be 
made  more  attractive  and  profitable  to  the  beginner.  The 
angles  and  numbers  in  these  problems  are  so  chosen  that  this 
table  contains  all  that  is  needed  for  their  solution. 

In  Chapter  VIII  complex  number  is  expressed  as  an  arith- 
metic multiple  of  a  quality  unit  in  its  trigonometric  type 
form,  and  the  fundamental  properties  of  such  number  are 
demonstrated.  The  proof  of  De  Moivre's  theorem  is  simple 
and  general,  and  its  meaning  and  use  are  fully  illustrated. 

Before  beginning  the  study  of  complex  number  one  should 
gain  a  clear  idea  of  the  sum,  or  resultant,  of  two  directed 
forces,  and  by  repeated  experiment  make  this  idea  familiar. 

It  is  believed  that  the  order  of  the  text  is  the  best  for  begin- 
ners ;  but,  with  the  exception  of  a  few  articles,  Chapter  I  may 
be  omitted  by  those  who  are  prepared  to  take  up  at  once  the 
general  treatment  in  Chapter  II.  Too  much  stress  cannot  be 
laid  on  careful  and  accurate  construction  and  measurement  in 
the  first  chapter.  Chapters  VII  and  VIII  and  the  latter  parts 
of  Chapter  VI  may  be  omitted  by  those  who  wish  a  shorter 
course. 

In  preparing  this  book  the  author  has  consulted  the  best 
authorities,  both  American  and  European.  Many  of  the  exam- 
ples have  been  taken  from  these  sources.  The  author  takes 
this  opportunity  to  express  to  many  teachers  and  other  friends 
his  appreciation  of  their  valuable  suggestions  and  their  en- 
couragement in  the  preparation  of  this  work. 

James  M.  Taylor. 

Colgate  University,  December,  1903. 


CONTENTS 


CHAPTER  I 

TRIGONOMETRIC  RATIOS  OF  ACUTE  ANGLES 

Section  Page 

1-3.    Trigonometric  ratios  defined,  one-valued 1, 2 

4.    Construction  of  angles  having  given  ratios 3, 4 

5, 6.    Approximate  values,  and  changes  from  0°  to  90°      .     .     .  5-7 

7-9.   Trigonometric  ratios  of  angles  of  triangle  and  of  co-angles  8, 9 

10,11.    Trigonometric  ratios  of  45°,  30°,  60° 10,11 

12,13.    Solving  right  triangles .  11-13 

14, 15.    Definitions  and  problems 14-21 


CHAPTER  II 

TRIGONOMETRIC  RATIOS  OF  POSITIVE  AND  NEGATIVE  ANGLES  OF  ANY  SIZE 

16.   Positive  and  negative  angles  of  any  size 22, 23 

17-19.    Coterminal  angles.     Quadrants 24,25 

20,21.    Trigonometric  ratios  of  any  angle 26,27 

22,23.    Laws  of  quality.     Sin-^c,  cos-^c,  ••  • 28-30 

24, 25.    Fundamental  identities.     Proof  of  identities 31-35 

26.  Changes  in  ratios  as  angle  changes  from  0°  to  360°  .     .     .  36-38 

27.  Trigonometric  ratios  of  0°,  90°,  180°,  270°  ......  39 

28, 29.    Trigonometric  ratios  of  -  ^  and  90°  +  ^       40-43 

30, 31.    Trigonometric  ratios  of  90°  -  A,  180°  -  ^,  n  90°  ±  ^  .     .  44, 45 

32,  33.   Trigonometric  lines  representing  the  ratios 46-51 

V 


vi  CONTENTS 


CHAPTER  III 


TRIGONOMETRIC  RATIOS  OF  TWO  ANGLES 

Section  Page 

34-37.    Trigonometric  ratios  of  the  sum  and  difference  of  two  angles  52-56 

38, 39.    Trigonometric  ratios  of  twice  and  half  an  angle  ....  57-59 

40.    Sums  and  differences  of  trigonometric  ratios 60-63 


CHAPTER  IV 

SOLUTION  OF  RIGHT  TRIANGLES  WITH  LOGARITHMS 

41-48.    Properties  and  computation  of  logarithms 64-68 

49.    Right-angled  triangles ".     .     69-71 

50,51.    Isosceles  triangles  and  regular  polygons 72,73 


CHAPTER  V 
SOLUTION  OF  TRIANGLES  IN  GENERAL 

52.    Law  of  sines  and  law  of  cosines 74,  75 

53-55.    The  four  cases  solved  without  logarithms 76,  77 

56.    Law  of  tangents 78, 79 

57-59.    First  three  cases  solved  with  logarithms 80-87 

60.  Trigonometric  ratios  of  half  angles  and  Case  (iv)  solved 

with  logarithms 88-91 

61.  Areas  of  triangles 92 

62-64.    Circumscribed,  inscribed,  escribed  circles 93, 94 


CONTENTS  vii 


CHAPTER  VI 


RADIAN  MEASURE,  GENERAL  VALUES,  TRIGONOMETRIC  EQUATIONS, 

INVERSE  FUNCTIONS 

Section  Paok 

65-67.    Radian  measure  of  angles 95-97 

68.   Principal  values 98 

69-71.    Angles  having  the  same  trigonometric  ratio     ....  99, 100 

72,  73.    Solution  of  trigonometric  equations 101-104 

74.    Trigonometric  functions        105 

75,  76.   Inverse  trigonometric  functions 106-109 


CHAPTER  VII 

PERIODS,  GRAPHS,  IMPORTANT  LIMITS,  COMPUTATION  OF  TABLE, 
HYPERBOLIC  FUNCTIONS 

77.  Periods  of  the  trigonometric  functions 110 

78-81.  Graphs  of  the  trigonometric  functions 111-114 

82.  Limit  of  the  ratio  of  sin  d  or  tan  ^  to  0 115 

83, 84.  Computation  of  trigonometric  functions 116, 117 

85,86.  Hyperbolic  functions 118,119 


CHAPTER  VIII 
COMPLEX  NUMBERS.     DE  MOIVRE'S  THEOREM 

87.    Quality  units  ±1,  ±  V^ 120 

88, 89.    Directed  lines  and  forces 121, 122 

90.    Complex  numbers 123,124 

91,92.    General  quality  unit.     Products  of  quality. units       .     .  125,126 

93.   De  Moivre's  theorem.     Quotients  of  quality  units    .     .  127, 128 


viii  CONTENTS 

Section  Page 

94, 95.  Products  and  quotients  of  complex  numbers    ....  128, 129 

96,  97.  The  gth  roots  of  cos  0  +  i  sin  0  and  (cos  0  +  i  sin  0)  r  .  130, 131 

98.  Exponential  form  for  cos  0  +  i  sin  0       132, 133 


CHAPTER  IX  ' 

MISCELLANEOUS  EXAMPLES 

Trigonometric  identities 134-136 

Trigonometric  equations  and  systems 137-141 

Problems  involving  triangles 142-146 

Problems  involving  areas  and  regular  polygons 146-147 

Formulas 148-151 


ANSWERS 153-171 


PLA^E   TRIGOI^OMETRY 

CHAPTER   I 
TRIGONOMETRIC  RATIOS  OF  ACUTE  ANGLES 

1.  Let  A  denote  the  number  of  degrees  in  the  acute  angle 
XOB]  then  A  is  the  numerical  measure,  or  measure,  of  this 
angle,  and  we  can  write  Z  XOB  =  A. 

From  any  point  in  either  side   of  the '  angle   XOB,  as  P, 
draw   PM  perpendicular    to    the 
other  side. 

Observe  that  0  is  the  vertex 
of  the  angle,  and  M  is  the  foot  of 
the  perpendicular  drawn  from  P. 
This  lettering  should  be  fixed  in     ''  no  i 

mind  so  that  in  the  following  defi- 
nitions the  lines  MP,  OM,  and  OP  shall  always  mean  the  same 
lines  as  in  fig.  1. 

The  six  simple  ratios  (three  ratios  and  their  reciprocals) 
which  can  be  formed  with  the  three  lines  MP,  OM,  OP  are 
called  the  trigonometric  ratios  of  the  angle  XOB,  or  A. 

These  ratios  are  named  as  follows : 

The  ratio  MP /OP    is  the  sine  of  A  ; 

and  its  recijirocal  OP /MP    is  the  cosecant  of  A. 

The  ratio  OM/OP    is  the  cosine  of  A  ; 

and  its  reciprocal  OP/OM    is  the  secant  of  A. 

The  ratio  MP/OM  is  the  tangent  of  A  ; 

and  its  reciprocal  OHL/IS.'P  is  the  cotangent  of  A, 
I 


PLANE  TRIGONOMETRY 


For  brevity  the  sine  of  A  is  written  sin  A ;  the  cosine  of  Aj 
cos  A;  the  tangent  of  A,  tan  A;  the  cotangent  of  A,  cot  A;  the 
secant  of  A,  sec  A ;  and  the  cosecant  of  A,  esc  A. 

Observe  that  sin  ^  is  a  compound  symbol  which,  taken  as  a 
whole,  denotes  a  number.     The  same  is  true  of  cos  A,  tan  A,  etc. 

Ex.  1.  What  four  trigonometric  ratios  of  the  angle  A  involve  the 
line  MP?  the  line  OM?  the  line  OP? 

Ex.  2.     What  trigonometric  ratios  are  reciprocals  of  each  other  ? 
Ex.  3.     Which  is  the  greater,  tan  A  or  sec  A?  cot  A  or  esc  A  ?    Why  ? 
Ex.  4.     Can  sin  A  or  cos  A  exceed  1  ?     Why  ? 

2.  An]/  trigonometric  ratio  of  a  given  angle  has  07ily  one 
value. 

Let  XOB  be  any  acute  angle.     Draw  PM_L  OX,  P'M'  ±  OX, 

P"M"  ±  OB ;  then,  by  §  1, 
^  .  MP    M'P' 

or  M"P"/OP".  (1) 

From  the  similarity  of  the  A 
[^//'^  OMP,  OM'P',   OM"P"  it  follows 
that  the  three  ratios  in  (1)  (or 
their  reciprocals)  are  all  equal; 
hence  sin  XOB  (or  esc  XOB)  has  but  one  value. 

Also,  from  the  similarity  of  these  A,  each  of  the  other  trigo- 
nometric ratios  of  Z  XOB  has  only  one  value. 

3.  Two  acute  angles  are  equal  if  any  trigonometric  ratio  of 
the  one  is  equal  to  the  same  ratio  of  the  other. 

Take  O^P^  in  fig.  3  equal  to  OP  in  fig.  2,  and  draw 
Pi  Ml  ±  OjXi-     We  are  to  prove  that 

if  sin  ZiOiPi  =  sin  XOP, 

i.e.  if  M,P^/OiPi  =  MP /OP,  (1) 

then  ZXiOiPi  =  ZXOP,  (2) 


TRIGONOMETRIC  RATIOS  OF  ACUTE  ANGLES 


By  construction,        O^Px  =  OP. 
Hence,  from  (1),       M^P^  =  MP. 

Therefore,  by  Geometry,  the  right  tri- 
angles OiPiMi  and  0PM  are  equal  in 
all  their  parts.  O^ 

Hence       Z  X^O^P^  =  Z  XOP. 

In  like  manner  the  student  should  prove  the  equality  of 
two  acute  angles  when  any  other  trigonometric  ratio  of  the 
one  is  equal  to  the  same  ratio  of  the  other. 

4.  Having  given  the  value  of  any  trigonometric  ratio  of  an 
oAiute  angle ^  to  construct  the  angle  and  obtain  the  values  of  its 
other  trigonometric  ratios. 

This  problem  will  be  illustrated  by  particular  examples. 

Ex.  1.  If  ^  is  an  acute  angle  and  cos^  =3/5,  construct  A  and  find 
the  values  of  its  other  trigonometric  ratios. 

Here         cos  A  =  DM/  OP  =  S/b. 
Hence,  if      OP  =  5  units,  DM  =  3  units. 

Let  0,  in  fig.  4,  be  the  vertex  of  the  angle 
A,  and  OX  one  of  its  sides. 

On  OX,  to  some  scale,  lay  off  OM  equal 
to  3  units,  and  at  M  draw  MS  ±  OX. 

With  O  as  a  center  and  with  a  radius 
equal  to  5  units,  draw  an  arc  cutting  MS  in 
some  point  as  P.     Draw  OPB. 

Then  ZXOB  =  A. 

For  cos  XOB  =  3  /  5  =  cos  ^. 

Hence,  by  §  3,    Z  XOB  =  A. 

Again,  MP  =  V52  -  32  units  =  4  units. 

Hence  sin  ^=4/5,  esc  ^  =  5/4 

cos^  =  3/5,  sec  ^=5/3; 

tan^  =  4/3,  cot  ^  =  3/4. 

Observe  that  6  is  the  numerical  measure  of  OP,  4  of  MP,  and  3  of  OM. 


Fig.  4 


51 


PLANE  TRIGONOMETRY 


Ex.  2.     If  A  is  an  acute  angle  and  sin  ^  =  2/3,  construct  A  and  find 
the  values  of  its  other  trigonometric  ratios. 

Here         sin  A  =  MP/  OP  =  2/3. 
Hence,  if     OP  =  3  units, 
MP  =  2  units. 

At  M,  in  fig.  5,  draw  MS  ±  OX  and 
lay  off  MP  equal  to  2  units. 

With  P  as  a  center  and  3  units  as  a 
radius,  strike  an  arc  cutting  OX  in  some 
point,  as  0.     Draw  OPB. 

Then  ZXOB  =  A. 

For  sinXO^  =  2/3  =  sin^. 

Hence,  by  §  3,  Z  XOB  =  A. 


M   X 


Fig.  5 


Again, 
Hence 


§3 


OM  =  V  32  -  22  units  =  V5  units, 
sin  J.  =  2/3,  CSC  ^=3/2; 

cos^=V5/3,  sec^  =  3/V5; 

tan^=2/V5,  cot^=V5/2. 

If  the  vertex  of  the  angle  A  were  required  to  be  at  some  fixed  point 
on  OX,  as  0,  we  would  draw  OK  ±  OX,  lay  off  OQ  equal  to  2  units, 
through  Q  draw  QN  parallel  to  OX,  and  with  0  as  a  center  and  3  units 
as  a  radius,  strike  an  arc  cutting  QN  at  P ;  then  draw  OPB  as  before. 

By  using  a  protractor,  i.  e.  a  graduated  semicircle,  we  find  that  Z  XOB, 
or  A,  is  an  angle  of  about  42°. 

Note.  The  student  should  form  the  habit  of  gaining  a  clear  idea  of  the 
size  of  an  angle  from  the  value  of  any  one  of  its  trigonometric  ratios. 


EXERCISE  I 

Construct  the  acute  angle  A,  obtain  the  values  of  all  its  trigonometric 
ratios,  and  find  its  size  in  degrees,  when  : 

1.  sin^=2/5.  6.  tan^  =  4/3. 

2.  sin^  =4/5.  7.  cot^  =5/2. 

3.  cos^=3/4.  8.  cot^=l/3. 

4.  cos^=l/3.  9.  sec^  =5/3. 

5.  tan^=l/4.  10.  sec^=4/3. 


11.  csc^  =5/2. 

12.  csc^  =3/2. 

13.  tan^  =  4,  or  4/1. 

14.  cot  J.  =  7,  or  7/1. 

15.  tan  A  =  9, 


TRIGONOMETRIC  RATIOS  OF  ACUTE  ANGLES 


16.    Express  each  of  the  trigonometric  ratios  of  an  acute  angle  A  in 
terms  of  its  sine,  writing  (sin^)^  in  the  form  sin^^. 
Infig.  1,  let  OP=l. 


Then 
Whence 

and 
Hence 


MP  /  OP  =  MP /I,  i.e.  sin  A  is  the  measure  of  MP. 
MP  =  sin  A, 


OM  =  ^OP^  -  MP^  =  Vl  -  sin2^. 

cos  A  =  OM/OP 
.-,  sec  A  = 

tan  A  = 
.-.  cot  A  =  Vl  —  sin2^/sin  A. 

CSC  A  =  OP /MP  =  1/sin  A. 


Vl  -  sin2^  ; 
1/Vl  -sin2^. 
MP/  OM  =  sin  A/Vl  -  sin'M 


§1 


17.  Express  each  of  the  trigonometric  ratios  of  an  acute  angle  in  terms 
of  its  cosine. 

In  fig.  1,  let  0P=1. 

Then  OM/  OP  =  OM/l,  i.e.  cos  A  is  the  measure  of  OM. 

Whence  OM  =  cos  A, 

and  MP  =  V^  -  OM^  =  Vl  -  cos^^,  etc. 

18.  Express  each  of  the  trigonometric  ratios  of  an  acute  angle  in  terms 
of  its  tangent. 

Infig.  1,  let  0M=  1. 

Then  MP/  OM  =  MP /I,  i.e.  tan  A  is  the  measure  of  MP. 

Whence  MP  =  tan  A, 

and  OP  =  Vo^^  +  I^  =  Vl  +  tan^^,  etc. 

5.    To  find  approximately  by  measurement  the  values  of  the 
trigonometric  ratios  of  any  given  angle. 

Ex.  1.  Find  by  construction  and 
measurement  the  values  of  the  six  trigo- 
nometric ratios  of  40°. 

With  a  protractor  lay  off  Z  XOB  =  40°. 

Take  OP  any  convenient  length,  say  10 
units  (the  longer  the  better),  and  draw 
PM  ±  OX.  By  careful  measurement  we 
find  that 

MP  =  6.4  units, 

OM  =  1.1  units,  approximately.  Fig.  6 


6  PLANE  trigonomj:try 

Hence,  as  approximate  values,  we  have 

sin  40°  =  6.4/10  =  0.64,  esc  40°  =  10/6.4  =  1.56; 

cos40°  =  7.7/10  =  0.77,  sec  40°  =  10/7.7  =  1.3; 

tan40°  =  6.4/7.7  =  0.83,  cot40°  =  7.7/6.4  =  1.2. 

Observe  that  instead  of  taking  OP  equal  to  10  units,  we  could  take 
OM  equal  to  10  units. 

Ex.  2.  By  construction  and  measurement  find  the  approximate  values 
of  the  sine,  cosine,  and  tangent  of  5°,  10°,  15°,  20°,  25°,  35°,  and  compare 
the  results  obtained  with  their  values  given  in  the  table  below. 

On  a  scale  of  20  to  an  inch  take  OM  equal  to  100  units  and  draw  MP 
±  OM.  With  O  as  their  common  vertex  and  OM  as  their  common  lower 
side,  draw  the  angles  5°,  10°,  etc. 

Observe  that  the  cosecant,  the  secant,  and  the  cotangent  of  any  angle 
can  be  obtained  by  taking  the  reciprocal  of  the  sine,  the  cosine,  and  the 
tangent  respectively  of  the  same  angle. 


Angle 

sin 

CSC 

cos 

sec 

tan 

cot 

1° 

0.0175 

57.2987 

0.9998 

1.0002 

0.0175 

57.2900 

5° 

0.0872 

11.4737 

0.9962 

1.0038 

0.0875 

11.4301 

10° 

0.1736 

5.7588 

0.9848 

1.0154 

0.1763 

5.6713 

15° 

0.2588 

3.8637 

0.9659 

1.0353 

0.2679 

3.7321 

20° 

0.3420 

2.9238 

0.9397 

1.0642 

0.3640 

2.7475 

25° 

0.4226 

2.3662 

0.9063 

1.1034 

0.4663 

2.1445 

30° 

0.5000 

2.0000 

0.8660 

1.1547 

0.5774 

1.7321 

35° 

0.5736 

1.7434 

0.8192 

1.2208 

0.7002 

1.4281 

40° 

0.6428 

1.5557 

0.7660 

1.3054 

0.8391 

1.1918 

45° 

0.7071 

1.4142 

0.7071 

1.4142 

1.0000 

1.0000 

50° 

0.7660 

1.3054 

0.6428 

1.5557 

1.1918 

0.8391 

55° 

0.8192 

1.2208 

0.5736 

1.7434 

1.4281 

0.7002 

60° 

0.8660 

1.1547 

0.5000 

2.0000 

1.7321 

0.5774 

65° 

0.9063 

1.1034 

0.4226 

2.3662 

2.1445 

0.4663 

70° 

0.9397 

1.0642 

0.3420 

2.9238 

2.7475 

0.3640 

75° 

0.9659 

1.0353 

0.2588 

3.8637 

3.7321 

0.2679 

80° 

0.9848 

1.0154 

0.1736 

5.7588 

5.6713 

0.1763 

85° 

0.9962 

1.0038 

0.0872 

11.4737 

11.4301 

0.0876 

89° 

0.9998 

1.0002 

0.0175 

57.2987 

57.2900 

0.0175 

In  the  above  table  observe  how  each  trigonometric  ratio  of  A  changes 
as  A  increases  from  1°  to  89°. 


TRIGONOMETRIC  RATIOS  OF  ACUTE  ANGLES 


6.    Changes  of  the  trigonometric  ratios  of  the  angle  A  as  A 
increases  from  0°  to  90°. 

In  fig.  7  conceive  the  line  OP  to  revolve  from  OX  to  OB,  i.e. 
suppose  Z  XOP,  or  A^  to  increase 
from  0°  to  90°. 

When  OP  coincides  with  OX, 

MP  =  0,  and  OM  =  OP. 

When  OP  coincides  with  05, 

MP  =  OP,  and  OM  =  0. 


O  M 


Hence  sin  0°  =  0,  sin  90°  =  1, 
cos  0°  =  1,  and  cos  90°  =  0. 

Therefore    when    A    increases 
from  0°  to  90°  sin  A  increases  from  0  to  1  and  cos  A  decreases 
from  1  to  0. 

Example.     What  is  the  vahie  of  tan  0°  ?  sec  0°  ?  cot  90°  ?  esc  90°  ? 

When  OP  is  very  near  to  OB,  OM  is  very  small ;  hence 
MP /  OM  is  very  large,  and  the  nearer  OP  is  to  OB,  the  larger 
is  MP  /  OM,  or  tan  .1.  So  that  as  A  approaches  90°,  tan  ^ 
increases  rapidly  and  can  exceed  any  constant  number  how- 
ever great ;  that  is,  tan  A  becomes  an  infinite  (denoted  by  oo). 
Hence  as  A  increases  from  0°  to  90°,  tan  A  increases  from  0 
to  CO ;  likewise  sec  A  increases  from  1  to  oo. 

Similarly,  when  OP  approaches  nearer  and  nearer  to  OX, 
OM / MP,  or  cot  A,  and  OP / MP,  or  esc  A,  become  infinites. 

Hence  as  the  angle  A  increases  from  0°  to  90°, 

sin  A  increases  from  0  to  1, 
cos  A  decreases  from  1  to  0, 
tan  A  increases  from  0  to  oo, 
cot  A  decreases  from  oo  to  0, 
sec  A  increases  from  1  to  oo, 
CSC  A  decreases  from  oo  to  1. 


PLANE  TRIGONOMETRY 


7.    Trigonometric  ratios  of  the  acute  angles  of  a  right  triangle. 

Let  A  and  B  be  the  measures  of 
the  acute  angles  of  the  right  tri- 
angle ABC,  a  the  measure  of  the 

ct  side  opposite  the  angle  A,  b  that  of 
the  side  opposite  B,  and  c  that  of 

G  the  hypotenuse  ;  then,  by  §  1,  we 
have 


Fig.  8 


sin  ^  =  -  =  -r ^ , 

c        hypotenuse 

CSC  A  =  -^ 
a 

b      side  adiacent 

cos  ^  =  -  =  — 7^* J 

c        hypotenuse 

sec  ^  =75 
b 

a      side  opposite 
0      side  adjacent 

b 
cot  A  =-' 
a 

Similarly  we  have 

sin  B  =  b/c,                 esc  5 

=  c/b', 

cos  B  =  a/c,                sec  5 

=  c/a-, 

t2in.B  =  b/a,                 cot  B 

=  a/b. 

8.    Complementary  angles.     Two  angles 

are  said  to  be  com 

plementary  when  their  sum  is  90°. 

E.g.,  the  complement  of  35°  is  90°  —  35°,  or  55° ;  the  complement  of 
70°  is  90°  -  70°,  or  20° ;  the  complement  of  A  is  90°  -  ^  ;  and  in  any 
right  triangle,  as  ABC  fig.  8,  the  acute  angles  A  and  B  are  complementary 
angles. 

9.    Trigonometric  ratios  of  complementary  angles,     li  Z.CAB 

=  A,  then  Z  CBA  =  90°  -  A.     Hence,  by  §  7,  we  have 

sin  (90°  —  A)=b/c  =  cos  A, 
cos  (90°  —  A)  =  a/c=  sin  A, 
tan  (90°  -A)  =  b/a  =  cot  A, 
cot  (90°  -  A)  =  a/b  =  tan  A, 
sec  (90°  —  A)  =  G/a  =  esc  A, 
G8C{90°  -  A)=c/b  =  sec  A. 


TRIGONOMETRIC  RATIOS  OF  ACUTE  ANGLES        9 

If  we  call  the  cosine  the  co-ratio  of  the  sine,  the  sine  the 
co-ratio  of  the  cosine,  the  cotangent  the  co-ratio  of  the  tangent, 
the  tangent  the  co-ratio  of  the  co- 
tangent, etc.,  then  the  six  identities 
above  can  be  summed  up  as  follows  : 

Any  trigonometric  ratio  of  an  acute 
angle  is  equal  to  the  co-ratio  of  its 
complement. 

E.g.,  since  60°  and  30°  are  comple- 
mentary angles,  we  have 

sin  60°  =  cos  30°,     tan  60°  =  cot  30°,     esc  60°  =  sec  30° 
Again,  since  45°  is  the  complement  of  itself,  we  have 

sin  45°  =  cos  45°,     tan  45°  =  cot  45°,     esc  45°  =  sec  45°. 
Ex.  1.     From  the  upper  half  of  the  table  in  §  5  obtain  the  lower  half. 
Since  50°  -f  40°  =  90°,  sin  50°  =  cos  40°  =  0. 7660  ; 

since  55°  +  35°  =  90°,  sin  55°  =  cos  35°  =  0.8192  ; 

since  60°  -f-  30°  =  90°,  sin  60°  =  cos  30°  =  0.8660  ;  etc. 

Ex.  2.     The  angle  A  being  acute,  find  the  value  of  A  in  the  equation 
sin^  =  cos2^.  (1) 

If  in  equation  (1)  we  substitute  for  cos  2  A  its  *  identical  expression 
sin  (90°  —  2  A),  by  Algebra  we  obtain  the  equivalent  equation  (2). 

*  Algebraic  definitions.  Two  numeral  expressions  which  denote  the  same 
number,  or  any  two  mathematical  expressions  which  denote  equal  numbers 
for  all  values  of  their  letters,  are  called  identical  expressions.  E.g.,  the  numeral 
expressions  4x3  and  8  +  4  are  identical,  so  also  are  the  literal  expressions 
(a  -\-b)(a  -  b)  and  a"^  -  62,  or  cos  A  and  sin  (90°  -  A). 

An  equality  is  the  statement  that  two  mathematical  expressions  denote  the 
same  number. 

An  equality  whose  members  are  identical  expressions  is  called  an  identity. 
An  identity  is  to  he  proved. 

An  equality  whose  members  are  not  identical  expressions  is  called  an 
equation.     An  equation  is  to  be  solved. 

The  sign  of  identity,  =,  read  "is  identical  with,"  is  often  used  instead  of 
the  sign  of  equality  =  in  writing  an  identity  whose  members  involve  one  or 
more  letters.  E.g.,  to  indicate  that  the  equality  sin^  =  cos  (90°  —  A)  is  an 
identity  and  not  an  equation  we  write  sin  A  =  cos  (90°  —  A) . 

Since  we  know  that  any  equality  which  involves  only  numerals  must  be  an 
identity,  the  sign  of  identity  is  used  only  in  writing  literal  identities. 


10 


PLANE  TRIGONOMETRY 


From  (2),  by  §  3, 


sin  A 
A 


sin  (90°  -  2  A). 
90°-  2  A,  ..A 


(2) 


Equation  (1)  is  a  trigonometric  equation,  and  the  only  value  of  the 
acute  angle  A  which  will  satisfy  it  is  30°. 


EXERCISE  n 

1.  By  §  9,  cos  30°  equals  what  ?   sin  60°?   cot  35°?   tan  15°?   sec  85°? 
CSC  76°  ?   sin  73°  14'  ?  cos  65°  43'  ? 

A  being  an  acute  angle,  find  its  value  in  each  of  the  following  equations: 

2.  sec  ^=  CSC  ^.  6.   sec  (75°  +  ^)  =  esc  2^. 

3.  tan  ^  =  cot  2  A.  7.   cot  {A  +  50°)  =  tan  7  A. 

4.  sin  2  ^  =  cos  3  A.  8.    sin  nA  =  cos  niA. 

5.  tan  ^  /2  =  cot  2  A.  9.   tan  cA  =  cot  (30°  -  A). 

10.    Trigonometric  ratios  of  45°.     Let  ABC  be  an  isosceles 
right  triangle  in  which 

AB  =  2  units, 
A=B  =  45°. 

AC  =  BC  =  V2  units, 
sin  45°  =  V2/2  =  cos  45°, 
tan  45°  =  1  =  cot  45°, 
sec  45°  =  V2  =  CSC  45°. 


Fig.  10 


11.  Trigonometric  ratios  of  30°  and 
60°.  Let  ABD  be  an  equilateral  tri- 
angle in  which  AB  =  2  units.  Draw 
BC±AD. 


Then 


AC  =  1, 
BC  =  ^S, 
A  =  60°, 
Z  ABC  =  30°. 


TRIGONOMETRIC  RATIOS  OF  ACUTE  ANGLES      11 


By§l,         sin  30°  =  1/2, 

CSC  30°  =  2 ; 

cos30°  =  V3/2, 

sec  30°  =  2/ V3; 

tan30°  =  l/V3, 

cot  30°  =  V3. 

Also,  by  §  1, 

sin  60°  =  V3/2, 

CSC  60°  =  2/ V3; 

cos  60°  =  1/2, 

sec  60°  =  2  ; 

tan  60°  =  V3, 

cot60°  =  l/V3,  or  V3/3. 

To  aid  the  memory  observe  that  sin  30°,  sin  45°,  and  sin  60°  are 
respectively  equal  to  Vl*  V2,  and  V3,  divided  by  2. 

It  is  easy  to  read  off  the  trigonometric  ratios  of  30°,  45°,  and  60°,  vsrhen 
we  keep  in  mind  the  figures  10  and  11. 

Example.  By  §  9  obtain  the  values  of  the  trigonometric  ratios  of  60° 
from  those  of  30°,  and  those  of  30°  from  those  of  60°. 

12.  Approximate  measurements  and  computations.  The  student 
should  remember  that  in  all  actual  measurements  the  results 
are  only  approximate.  It  is  impossible  to  measure  any 
quantity  with  absolute  accuracy.  The  degree  of  accuracy 
sought  will  depend  upon  the  importance  of  the  results.  The 
degree  of  accuracy  secured  will  depend  upon  the  instruments, 
methods,  and  care  which  are  used.  Likewise,  in  practical 
computations,  a  sum,  difference,  product,  or  quotient  of  two 
approximate  values  will  not  have  a  greater  degree  of  accuracy 
than  that  of  the  least  accurate  of  the  two  values.  E.g.,  if  one 
numerical  measure  is  accurate  to  two  figures  and  another  to 
three  figures,  their  sum,  product,  or  quotient  will  not  in  general 
be  accurate  to  more  than  two  figures.  If  each  of  two  numerical 
measures  has  three-figure  accuracy,  or  if  one  has  four-figure 
accuracy  and  the  other  only  three-figure  accuracy,  their  product 
or  quotient  will  not  in  general  have  more  than  three-figure 
accuracy.  The  values  tabulated  in  §  5  have  only  four-figure 
accuracy. 


12  PLANE  TRIGONOMETRY 

13.  Solving  right  triangles.  Of  the  six  parts  (three  sides 
and  three  angles)  of  a  right  triangle,  one  part  (the  right 
angle)  is  always  known.  If,  of  any  right  triangle,  two  other 
parts  are  given  (one  at  least  being  a  side),  Geometry  proves 
that  the  triangle  is  entirely  determined,  and  shows  how  to 
construct  it. 

Trigonometry  shows  how  to  compute  the  numerical  values  of 
the  unknown  parts  of  a  triangle  when  the  known  parts  are 
sufficient  to  determine  it. 

This  process  is  called  solving  the  triangle. 

Hence,  in  solving  right  triangles,  we  must  consider  the  two 
following  cases : 

(i)  Given  one  side  and  one  acute  angle. 
(ii)  Given  any  two  sides. 

Case  (1).  Ex.  1.  In  the  right  triangle  ^^C,  ^  =  35°  and  BC  =  20  feet ; 
find  the  numerical  values  of  the  other  parts. 

Construction.     To  some  scale  (as  30  ft.  to  an 
inch)  construct  as  accurately  as  possible  a  right 
a    triangle  having  the   given  parts  A  =  35°  and 


Solution.     B  =  90°  -  A=90°  -Sb°=  65°. 

Now,  the  ratio  of  either  of  the  unknown  sides 
Fig.  12  ,       '  . ,         •        X  •  X  •        .•       J^ 

to  the  known  side  a  is  a  trigonometric  ratio  of 

35°,  and  any  trigonometric  ratio  of  35°  can  be  obtained  from  the  table 

in  §  5. 

Thus,  b/a  =  cot  ^  =  cot  35°.  §  1 

.-.  6/20  =  cot  35°  =  1.4281.  by  table 

.-.  &=:  1.4281  X  20=1:28.562. 
Again,  c/20  =  esc  35°  =  1.7434.  §  1,  table 

.•.c  =  1.7434  X  20  =  34.868. 

If  we  regard  20  as  exact,  or  at  least  accurate  to  four  figures,  the  values 
of  6  and  c  are  accurate  to  only  four  figures ;  for  cot  35°  and  esc  35°  are 
accurate  to  only  four  figures  (§12). 


SOLUTION  OF  TRIANGLES  13 

Check.     Measure  AC  and  AB,  and  multiply  the  number  of  inches  in 
each  by  the  number  of  feet  which  an  inch  represents.     We  thus  obtain 
h  =  28.6,  c  =  34.9. 

As  a  numerical  check  we  could  use 

a2  =  c2  -  62^  or  cfi  =  (c  +  6)  (c  -  h). 
But  for  simplicity  and  to  emphasize   the 
importance  of  accurate  construction,  we  shall, 
in  this  chapter^  use  only  the  check  by  construc- 
tion and  measurement. 

Fig.  13 
Case  (ii).     Ex.  2.     In    the    right    triangle 

ABC,  BC  =  83.91  ft.  and  AC  =  100  ft.  ;  find  the  other  parts. 

r^  =  4o°, 

Given  |"  ~  ^'^•^^'    to  find     \  B  =  50°, 

''='''■'  [C=1S0.64. 

Construct  the  triangle  ABC  having  the  given  parts. 

rtan^  =  a/6.  (1) 

Formulas  J         ^  =  90°  -  ^.  (2) 

[     c/6  =  sec^.  (3) 

Computation.    From  (1),  tan  A  =  83.91  / 100 

=  0.8391  =  tan  40°.  by  table 

Hence,  by  §  3,  A=  40°. 

From  (2),  B  =  90°  -  40°  =  50°. 

From  (3),  c  =  100  •  sec  40° 

=  100  X  1.3054  =  130.54.  by  table 

Check.     By  measurement  A  -  40°,  B  =  50°,  AB  =  ISl  ft.  nearly. 

The  solution  above  illustrates  the  five  steps  which,  in  the 
first  solutions  at  least,  should  be  kept  separate  and  distinct. 

(i)  Statement  of  the  problem, 

(ii)  Construction  of  the  triangle, 

(iii)  Writing  the  needed  formulas. 

(iv)  Making  the  computations. 

(v)  Applying  some  check  or  test  to  answers, 


14  PLANE  TRIGONOMETRY 

EXERCISE  in 

Solve  the  right  triangle  ABC^  when  : 


1. 

A  =  25°, 

a  =  30. 

9. 

6  =  93.97, 

c  =  100. 

2. 

B  =  55°, 

&  =  10. 

10. 

a  =17.1, 

c  =  50. 

3. 

A  =  65°, 

c  =  70. 

11. 

B  =  75°, 

c  =  40. 

4. 

B  =  15°, 

6  =  20. 

12. 

A  =  10°, 

6  =  30. 

5. 

B  =  35°, 

a  =  50. 

13. 

A  =  20°, 

c  =  80. 

6. 

B  =  55°, 

c  =  60. 

14. 

B  =  25°, 

a  =  30. 

7. 

a  =  36.4, 

b  =  100. 

15. 

a  =  30.21, 

c  =  33^. 

8. 

a  =  23.315, 

6  =  50. 

16. 

a  =  13.4, 

6  =  50. 

14.  A  vertical  line  at  any  point  is  the  line  determined  by 
the  plumb  line  at  that  point. 

A  horizontal  line  (or  plane)  at  any  point  is  the  line  (or  plane) 
which  is  perpendicular  to  the  vertical  line  at  that  point. 

A  horizontal  angle  is  an  angle  whose  sides  are  perpendicular 
to  the  vertical  line  at  its  vertex. 

A  vertical  angle  is  an  angle  whose  plane  contains  the  vertical 
line  at  its  vertex. 

A  vertical  angle  of  which  one  side  is  horizontal  is  called  an 
angle  of  elevation  or  an  angle  of  depression,  according  as  the 
second  side  is  above  or'  below  the  horizontal  side. 

Note.  All  vertical  lines  converge  towards  the  center  of  the  earth. 
But  in  the  next  two  definitions  any  two  vertical  lines  are  regarded  as 
parallel.  This  is  approximately  true  for  short  distances  and  is  always 
assumed  as  true  for  such  distances  unless  very  great  accuracy  is  required. 

The  horizontal  distance  between  two  points  is  the  distance 
from  one  of  the  two  points  to  a  vertical  line  through  the  other. 

The  vertical  distance  between  two  points  is  the  distance 
from  one  of  the  two  points  to  the  horizontal  plane  through 
the  other. 


PROBLEMS 


15 


Fig.  14 


E.g.^  let  MP  be  the  vertical  line  at  P  and  let  the  horizontal  plane  at 
A  cut  this  vertical  line  in  M ;  then  AM  is  called  the  horizontal  distance, 
and  MP  is  called  the  vertical 
distance,  between  the  points 
A  and  P.  Moreover  Z  MAP 
is  the  angle  of  elevation  of  P, 
as  seen  from  A.  Also,  if  PN 
is  horizontal  at  P  and  in  the 
plane  AMP,  Z  NPA  is  the 
angle  of  depression  of  A,  as 
seen  from  P. 

Now  if  we  assume  that  the  vertical  lines  at  A  and  P  are  parallel,  the 
lines  AM  and  NP  will  be  parallel  also  and  the  angles  MAP  and  NPA 
will  be  equal. 

15.  Solving  problems.  The  practical  problems  which  follow 
will  illustrate  the  utility  of  the  trigonometric  ratios  of  angles 
in  computing  heights,  distances,  angles,  areas,  etc. 

In  solving  problems  it  will  be  helpful  to  observe  the  fol- 
lowing general  method  of  procedure. 

First  step.  Construct  accurately  to  some  convenient  scale 
a  drawing  which  will  show  the  relations  of  the  given  angles 
and  lines  to  those  which  are  required. 

Second  step.  Draw  any  auxiliary  lines  which  may  be  help- 
ful in  the  trigonometric  solution.  By  examining  the  drawing, 
fix  upon  the  simplest  steps  which  are  necessary  to  solve  the 
problem. 

Third  step.  Write  the  needed  for- 
mulas. Make  the  computations,  and 
check  the  answers. 

Ex.  1.  A  man,  standing  on  the  bank  of  a 
river  at  P,  wishes  to  find  how  far  he  is  from 
a  tree  at  T  on  the  opposite  bank.  He  locates 
a  staff  at  S  so  that  PS±PT.  By  measure- 
ment he  finds  that  the  horizontal  distance 
PS  =  250  ft.,  and  that  the  horizontal  angle 
P-Sr=  40<^.     Find  the  distance  PT.  Fig.  15 


16 


PLANE  TRIGONOMETRY 


By  §  1,  FT / PS  =  tan  40°  =  0.8391.  by  table 

.:  PT=  250  ft.  X  0.8391  =  209.77  ft. 
Check.     By  measurement  PT  =  210  ft.  nearly,  when  PS  =  250  ft. 

Ex.  2.  A  vertical  flagstaff  stands  on  a  horizontal  plane.  At  a  point 
200  ft.  from  the  foot  of  the  staff  the  angle  of  elevation  of  its  top  is  found 
to  be  20°.     Find  the  height  of  the  flagstaff. 

Let  MP  (fig.  14)  represent  the  flagstaff,  and  A  the  point  from  which 
the  angle  of  elevation  is  taken. 

Then  ^Jlf=200ft., 

and  ZMAP  =  20°. 

By  §  1,  MP /AM  =  tan  20°  =  0.364.  by  table 

.-.  MP  =  200  ft.  X  0.364  =  72.8  ft. 
Check.     By  measurement  MP  =  73  ft.  nearly,  when  AM  =  200  ft. 

Ex.  3.  A  man  wishes  to  find  the  height  of  a  tower  DB  which  stands 
on  a  horizontal  plane.     From  a  point  A  on  this  plane  he  finds  the  angle 

^  of  elevation  of  the  top  of  the  tower  to 
be  35°.  From  aj  point  C,  which  is  in 
the  horizontal  plane  at  A  and  100  ft. 
nearer  the  tower,  he  finds  the  angle  of 
elevation  to  be  65°.  Find  the. height 
of  the  tower. 

Solution  1.     Let         DB  =  y  ft. 
Then     AD  /  y  =  cot  35°  =  1 .  4281 , 
Fig.  16  and  CD/y  =  cot  65°  =  0.4663. 

.-.  100  =  AD-  CD 

=  (1.4281  -  0.4663)  y. 
.:  y  =  103.97. 

Solution  2.  From  C  draw  CE  ±AB,  thus  forming  the  right  triangles 
^CJ^and  CEB. 

Now  Z  CBE  =  Z  ABD  -  Z  CBD 

=  55°  -  25°  =  30°. 
Let  CE  =  z  ft.  and  CB  =  w  ft. 

Then  ;2  / 100  =  sin  35°  =  0. 5736,  (1) 

w/z  =  cscS0°  =  2,  (2) 

and  y/w  =  sin  65°  =  0. 9063.  (3) 


PROBLEMS  17 

Multiplying  together  (1),  (2),  and  (3),  member  by  member,  we  obtain 

y/100  =  1.1472  X  0.9063,  or  y  =  103.97. 
Check.     By  measurement  DB  =  104  ft.  nearly,  when  ^C  =  100  ft. 

Ex.  4.     From  the  top  of  a  hill  300  ft.  higher  than  the  foot  of  a  tree, 
the  angles  of  depression  of  the  top  and  the  foot  of  the  tree  are  found  to 

be  20°  and  25°  respectively.     Find  ^ p 

the  height  of  the  tree. 

Let  P  be  the  top  of  the  hill,  B 
the  foot  of  the  tree,  and  C  its  top. 
In  the  plane  PBC  draw  PA  hori- 
zontal at  P  and  prolong  it  until  it 
intersects  the  vertical  line  BC  pro- 


ducedm  A. 

Fig.  17 

Then 

Z  APC  =  20°  and  Z  APB  =  25°. 

Let 

BC  =  x  ft. 

Then 

AC=  (300 -X)  ft. 

Hence 

^P/300  =  cot  25°  =  2.1445, 

(1) 

and 

(300 

-x)/AP  =  tan  20°  =  0.3640. 

(2) 

Multiplying  together  (1)  and  (2),  member  by  member,  we  obtain 
(300  -  a;)/300  =  2.1445  x  0.3640. 

.-.  X  =  65.82  nearly.  ' 

Check.     By  measurement  BC  =  66  ft.  nearly,  when  BA  =  300  ft. 

Note.  All  the  problems  in  the  following  exercise  need  not  be  solved 
before  beginning  Chapter  II.  The  solution  of  a  few  problems  at  a  time, 
while  the  student  is  pursuing  the  more  abstract  and  theoretic  portions 
of  the  science,  will  serve  to  keep  before  him  its  practical  utility  and 
maintain  his  interest. 

EXERCISE  IV 

1.  The  length  of  a  kite  string  is  250  yds.  and  the  angle  of  elevation 
of  the  kite  is  40°.  Find  the  height  of  the  kite,  supposing  the  line  of  the 
kite  string  to  be  straight.  Ans.    160.7  yds. 

2.  A  stick  10  ft.  in  length  stands  vertically  in  a  horizontal  area, 
and  the  length  of  its  shadow  is  8.391  ft.  Find  the  angle  of  elevation  of 
the  sun.  Ans.   50°. 


18  PLANE  TRIGONOMETRY 

3.  A  tree  is  broken  by  the  wind  so  that  its  two  parts  form  with  the 
ground  a  right-angled  triangle.  The  upper  part  makes  an  angle  of  35° 
with  the  ground,  and  the  distance  on  the  ground  from  the  trunk  to  the  top 
of  the  tree  is  60  ft.     Find  the  length  of  the  tree.  Ans.    96.06  ft. 

4.  The  distance  between  two  towers  on  a  horizontal  plane  is  60  ft. ,  and 
the  angle  of  depression  of  the  top  of  the  first  as  seen  from  the  top  of  the 
second,  which  is  150  ft,  high,  is  25°.     Find  the  height  of  the  first  tower. 

5.  At  a  point  200  ft.  from  the  base  of  an  unfinished  tower,  the  angle 
of  elevation  of  its  top  is  20° ;  when  completed,  the  angle  of  elevation  of 
its  top  at  this  point  will  be  30°.  How  much  higher  is  the  tower  to 
be  built? 

6.  The  angle  of  elevation  of  the  sun  is  65°  and  the  length  of  a  tree's 
shadow  on  a  level  plane  is  50  ft.     Find  the  height  of  the  tree. 

7.  A  chimney  stands  on  a  horizontal  plane.  At  one  point  in  this 
plane  the  angle  of  elevation  of  tlie  top  of  the  chimney  is  30°,  at  another 
point  100  feet  nearer  the  base  of  the  chimney  the  angle  of  elevation  of 
the  top  is  45°.     Find  the  height  of  the  chimney. 

8.  A  person  standing  on  the  bank  of  a  river  observes  that  the  angle 
subtended  by  a  tree  on  the  opposite  bank  is  50°  ;  walking  40  ft.  from  the 
bank  he  finds  the  angle  to  be  30°.  Find  the  height  of  the  tree  and 
the  breadth  of  the  river,  if  the  two  points  of  observation  are  in  the  same 
horizontal  line  at  the  base  of  the  tree. 

9.  The  shadow  of  a  tower  standing  on  a  horizontal  plane  is  found  to 
be  60  ft.  longer  when  the  sun's  altitude  is  30°  than  when  it  is  45°.  Find 
the  height  of  the  tower. 

10.  At  a  point  midway  between  two  towers  on  a  horizontal  plane  the 
angles  of  elevations  of  their  tops  are  30°  and  60°  respectively.  Show 
that  one  tower  is  three  times  as  high  as  the  other. 

11.  Two  observers  on  the  same  horizontal  line  and  in  the  same  vertical 
plane  with  a  balloon,  on  opposite  sides  of  it  and  2500  ft.  apart,  find  its 
angles  of  elevation  to  be  35°  and  55°  respectively.  Find  the  height  of  the 
balloon. 

12.  A  man  in  a  balloon  observes  that  the  bases  of  two  towers,  which 
are  a  mile  apart  on  a  horizontal  plane,  subtend  an  angle  of  70°.  If  he  is 
exactly  above  the  middle  point  between  the  towers,  find  the  height  of 
the  balloon. 


PROBLEMS  19 

13.  From  the  foot  of  a  tower  the  elevation  of  the  top  of  a  church 
spire  is  55°,  and  from  the  top  of  the  tower,  which  is  50  ft.  high,  the 
elevation  is  35°.  Find  the  height  of  the  spire  and  the  distance  of  the 
church  from  the  tower,  if  both  stand  on  the  same  horizontal  plane. 

14.  From  the  top  of  a  tower  whose  height  is  108  ft.  the  angles  of 
depression  of  the  top  and  bottom  of  a  vertical  column  standing  on  a  level 
with  the  base  of  the  tower  are  found  to  be  25°  and  35°  respectively. 
Find  the  height  of  the  column  and  its  distance  from  the  tower. 

15.  Two  pillars  of  equal  height  stand  on  opposite  sides  of  a  horizontal 
roadway  which  is  100  ft.  wide.  At  a  point  in  the  roadway  between  the 
pillars  the  angles  of  elevation  of  their  tops  are  50°  and  25°  respectively. 
Find  the  height  of  the  pillars  and  the  position  of  the  point  of  observation. 

16.  A  house  50  ft.  high  and  a  tower  stand  on  the  same  horizontal 
plane.  The  angle  of  elevation  of  the  top  of  the  tower  at  the  top  of  the 
house  is  25°,  on  the  ground  it  is  55°.  Find  the  height  of  the  tower  and 
its  distance  from  the  house. 

17.  On  the  top  of  a  bluff  is  a  tower  75  ft.  high  ;  from  a  boat  on  the 
bay  the  angles  of  elevation  of  the  top  and  base  of  the  tower  are  observed 
to  be  25°  and  15°  respectively.  Find  the  horizontal  distance  of  the  boat 
from  the  tower,  also  the  distance  of  the  boat  from  the  top  of  the  tower. 

18.  One  of  the  equal  sides  of  an  isosceles  triangle  is  50  ft.  and  one  of 
its  equal  angles  is  40°.  Find  the  base,  the  altitude,  and  the  area  of  the 
triangle. 

19.  The  base  of  an  isosceles  triangle  is  68.4  ft.  and  each  of  its  equal 
sides  is  100  ft.     Find  the  angles,  the  height,  and  the  area. 

20.  The  base  of  an  isosceles  triangle  is  100  ft.  and  its  height  is  35.01  ft. 
Find  its  equal  sides  and  the  angles. 

21.  The  base  of  an  isosceles  triangle  is  88  ft.  and  its  vertical  angle  is 
70°.     Find  the  height,  the  equal  sides,  and  area. 

22.  The  base  of  an  isosceles  triangle  is  100  ft.  and  the  equal  angles 
are  each  65°.     Find  the  equal  sides,  the  height,  and  the  area. 

23.  The  height  of  an  isosceles  triangle  is  60  ft.  and  its  vertical  angle 
is  30°.     Find  the  sides  and  the  area. 


20  PLANE   TRIGONOMETRY 

24.  A  man's  eye  is  on  a  level  with  and  100  ft.  distant  from  the  foot 
of  a  flag  pole  36.4  ft.  high.  When  he  is  looking  at  the  top  of  the  pole, 
what  angle  does  his  line  of  sight  make  with  a  line  from  his  eye  to  the 
foot  of  the  pole  ? 

25.  A  circular  pond  has  a  pole  standing  vertically  at  its  center  and  its 
top  is  100  ft.  above  the  surface.  At  a  point  in  the  circumference  the 
angle  subtended  by  the  pole  is  20°.  Find  the  radius  and  the  area  of  the 
pond. 

26.  A  ladder  33^  ft.  long  leans  against  a  house  and  reaches  to  a  point 
30.21  ft.  from  the  ground.  Find  the  angle  between  the  ladder  and  the 
house  and  the  distance  the  foot  of  the  ladder  is  from  the  house. 

27.  From  the  summit  of  a  hill  there  are  observed  two  consecutive 
milestones  on  a  straight  horizontal  road  running  from  the  base  of  the 
hill.  The  angles  of  depression  are  found  to  be  10°  and  5°  respectively. 
Find  the  height  of  the  hill. 

28.  At  the  foot  of  a  hill  the  angle  of  elevation  of  its  summit  is  observed 
to  be  30°;  after  ascending  the  hill  500  ft.,  up  a  slope  of  20°  inclination, 
the  angle  of  elevation  of  its  summit  is  found  to  be  40°.  Find  the  height 
of  the  hill  if  the  two  points  of  observation  and  the  summit  are  in  the  same 
vertical  plane. 

One  method  of  solution  is  similar  to  that  of  the  second  solution  of 
example  3  in  §  15. 

29.  At  the  foot  of  a  mountain  the  angle  of  elevation  of  its  summit  is 
35°;  after  ascending  an  opposite  mountain  3000  ft,,  up  a  slope  of  15° 
inclination,  the  angle  of  elevation  of  the  summit  is  15°.  Find  the  height 
of  the  first  mountain  if  the  points  of  observation  and  the  summit  are  in 
the  same  vertical  plane. 

30.  From  the  extremities  of  a  ship  500  ft.  long  the  angles  which  the 
direction  of  a  buoy  makes  with  that  of  the  ship  are  60°  and  75°.  Find  the 
distance  of  the  buoy  from  the  ship,  having  given  that  cot  76°  =  2  —  y/3. 

Ans.    125(V3  +  3)ft. 

31.  There  are  two  posts  which  are  240  and  80  ft.  high  respectively. 
From  the  foot  of  the  second  the  elevation  of  the  top  of  the  first  is  found 
to  be  60°.     Find  the  elevation  of  the  second  from  the  foot  of  the  first. 

Am.  30°. 


PROBLEMS  21 

32.  A  boy  standing  c  feet  behind  and  opposite  the  middle  of  a  football 
goal  sees  that  the  angle,  of  elevation  of  the  nearer  crossbar  is  A,  and  the 
angle  of  elevation  of  the  farther  one  is  B.  Show  that  the  length  of  the 
field  is  c  (tan  AcotB  —  1). 

33.  A  valley  is  crossed  by  a  horizontal  bridge  whose  length  is  I.  The 
sides  of  the  valley  make  angles  A  and  B  with  the  horizon.  Show  that 
the  height  of  the  bridge  above  the  bottom  of  the  valley  is  I /{cot  A  +  cot  B). 

34.  Two  forces  of  a  and  b  lbs.  respectively  act  in  the  same  direction. 
Find  their  resultant.     Illustrate  the  problem  geometrically. 

35.  Two  forces  of  a  and  b  lbs.  respectively  act  in  opposite  directions. 
Find  their  resultant  when  a > 6,  when  a  =  b,  and  when  a<b.  Illustrate 
each  case  geometrically. 

36.  By  two  or  more  experiments  verify  that,  if  in  any  triangle  ABC 
the  two  sides  AB  and  BC  represent  two  forces  (both  in  size  and  direc- 
tion), the  third  side  AC  will  represent  their  resultant,  i.e.  their  sum  in  its 
simplest  form. 

37.  Two  forces  of  3  and  4  lbs.  respectively  act  at  right  angles  to  each 
other.  Show  that  their  resultant  is  a  force  of  5  lbs.  and  that  its  line  of 
action  and  that  of  the  first  force  make  an  angle  whose  tangent  is  4/3. 

Suggestion.  In  fig.  4  let  OM  and  MP  respectively  represent  the  two 
forces,  then  the  line  OP  will  represent  the  resultant. 

38.  Two  forces  of  a  and  b  lbs.  respectively  act  at  right  angles.  Show 
that  their  resultant  is  a  force  of  Va^  +  6^  lbs. ,  and  that  its  line  of  action 
and  that  of  the  first  force  make  an  angle  whose  tangent  is  b/a. 

39.  Two  forces  act  at  right  angles.  The  first  is  a  force  of  3  lbs.  and 
the  resultant  is  one  of  5  lbs.  Show  that  the  second  force  is  one  of  4  lbs. , 
and  that  the  lines  of  action  of  the  first  force  and  the  resultant  form  an 
angle  whose  cosine  is  3/5. 

40.  Two  forces  act  at  right  angles.  The  first  is  a  force  of  a  lbs,  and  the 
resultant  is  one  of  c  lbs.  Show  that  the  second  force  is  one  of  Vc^  — a^  lbs. 
and  that  the  lines  of  action  of  the  first  force  and  the  resultant  form  an 
angle  whose  cosine  is  a/c. 


CHAPTEK   II 


TRIGONOMETRIC  RATIOS  OF  POSITIVE  AND  NEGATIVE 
ANGLES   OF  ANY  SIZE 

16.  Positive  and  negative  angles  of  any  size.  In  the  first 
chapter  we  studied  acute  angles  and  considered  their  size  only. 
When,  however,  we  conceive  an  angle  as  generated  by  a  rotat- 
ing line,  we  see  that  it  can  be  either  positive  or  negative  and 
of  any  size  whatever. 

Thus,  suppose  a  line  OP  to  start  from  OX  and  to  rotate 

about  0  counter-clockwise; 
that  is,  in  a  direction  oppo- 
site to  that  of  the  hands  of 
a  clock. 

When  OP  reaches  the 
position  OPi  it  has  gener- 
ated the  acute  angle  XOP^. 
When  OP  reaches  the  posi- 
tion OP2  it  has  generated  the 
obtuse  angle  XOP^.  When 
OP  reaches  OP^  it  has  gener- 
ated the  angle  XOP^  {i.e.  Z.  XOP^  + /.  P^OP^).  When  OP 
reaches  OP4  it  has  generated  the  angle  XOP^  (i.e.  Z  XOP^ 
H-  Z  P3OP4).  When  OP  reaches  OX  it  has  generated  an  angle 
of  360°. 

If  OP  continues  to  rotate,  when  it  reaches  OPi  the  second 
time  it  has  generated  the  angle  360°  -f-  the  acute  angle  XOP^ ; 
when  OP  reaches  OPi  the  third  time  it  has  generated  the  angle 
720°  H-  the  acute  angle  XOPi ;  and  so  on  for  any  number  of 
revolutions. 


TRIGONOMETRIC  RATIOS  28 

When  the  rotation  of  OP  is  counter-clockwise,  the  angle 
generated  is  said  to  be  positive ;  hence,  when  the  rotation  of 
OP  is  clockwise,  the  angle  generated  is  negative. 

E.g.,  in  ten  minutes  the  minute  hand  of  a  clock  generates  a  negative 
angle  of  00° ; 

in  15    minutes  it  generates  an  angle  of  —    90° ; 

u  30  it        u  ^i  ii       ii      u  _  180°: 

"    1    hour       "         "         "      ''      "  -  360^ 

"    3|  hours       "         "  "       "      "  -(3  X  360° +180°); 

and  so  on.     If  the  hands  of  a  clock  were  to  rotate  in  the  opposite  direction, 
i.e.  counter-clockwise,  they  would  genersite  positive  angles. 

The  line  OX  which  marks  the  first  position  of  the  rotat- 
ing line  OP  is  called  the  initial  side  of  the  angle  XOP^ ;  and 
the  line  OP3  which  marks  the  final  position  of  OP  is  called  the 
terminal  side  of  this  angle. 

The  size  of  an  angle  gives  the  amount  which  its  generating 
line  has  rotated,  and  its  quality  *  gives  the  direction  of  this 
rotation. 

The  value  of  a  positive  or  a  negative  angle  includes  both  its 
size  and  its  quality  as  positive  or  negative. 

17.  Coterminal  angles.  Any  angle,  positive  or  negative, 
which  has  the  same  initial  side  and  the  same  terminal  side 
as  angle  A  is  said  to  be  coterminal  with  A.  If  any  angle,  as 
XOP2  in  fig.  18,  is  increased  or  diminished  by  360°  (or  by  any 
entire  multiple  of  360°),  the  resulting  angle,  whether  positive 
or  negative,  will  have  the  same  initial  and  the  same  terminal 
side  as  XOP^. 

*  In  Algebra  the  quality  of  a  particular  number  as  positive  or  negative  is 
denoted  by  the  sign  +  or  — ,  and  this  quality  is  often  called  the  sign  of  the 
number.  It  is  unfortunate,  however,  to  use  the  same  word  sign  as  the  name 
both  of  a  symbol  and  also  of  the  property  of  number  denoted  by  this  symbol. 
Moreover  the  introduction  of  the  word  sine  adds  another  reason  for  not  calling 
the  quality  of  a  number  its  sign  in  Trigonometry. 


24  PLANE  TRIGONOMETKY 

Hence  if  n  is  any  integer,  positive  or  negative,  then  all  the 
angles,  and  only  those,  which  are  or  can  be  made  coterminal 
with  any  angle  A  are  denoted  by  n  360°  +  A.  E.g.,  2  -360°  +  40° 
is  or  can  be  made  coterminal  with  40°. 

Evidently  there  are  as  many  different  angles  coterminal 
with  A  as  there  are  different  entire  values  for  n. 

18.  Quadrants.  If,  as  in  fig.  18,  the  initial  side  of  the  angle 
XOP^  is  produced  through  the  vertex  to  X',  and  the  perpen- 
dicular YOY'  drawn  through  the  vertex  0,  these  lines  will 
divide  the  plane  of  the  angle  into  four  equal  parts  called 
quadrants.  These  quadrants  are  numbered  in  the  positive 
dii^ection,  reckoning  from  the  initial  side  of  the  angle  under 
consideration  ;  that  is,  if  OX  is  the  ifiitial  side  of  the  angle 
considered,  then  XOY  will  be  the  first  quadrant;  YOX'  the 
second  quadrant;  X'OY'  the  third  quadrant;  and  Y'OX  the 
fourth  quadrant. 

If  OF  is  the  initial  side  of  the  angle  considered,  then  YOX' 
will  be  the  first  quadrant ;  and  so  on. 

For  convenience,  an  angle  is  said  to  be  in  (or  of)  that 
quadrant  in  which  its  terminal  side  lies. 

E.g.,  the  angle  XOP2  (fig.  18)  is  said  to  be  in  the  second  quadrant, 
since  its  terminal  side  OP2  lies  in  that  quadrant ;  the  angle  XOP4  is  said 
to  be  in  the  fourth  quadrant,  since  its  terminal  side  OP4  is  in  that  quad- 
rant. The  angle  YOP2  is  in  the  first  quadrant,  and  YOP3  is  in  the 
second  quadrant,  since  here  OY  is  the  initial  side  and  YOX'  is  the  first 
quadrant. 

Again,  200°  =:  180°  +  20°,  hence  an  angle  of  200°  is  in  the  third  quad- 
rant ;  880°  =  2  (360°)  -f-  160°,  hence  an  angle  of  880°  is  in  the  second 
quadrant.  An  angle  of  —  50°  is  in  the  fourth  quadrant,  and  an  angle  of 
-  330°  is  in  the  first  quadrant.  Since  -  400°  =  -  360°  -  40°,  an  angle 
of  —  400°  is  in  the  fourth  quadrant. 

19.  Two  angles  are  said  to  be  complementary  when  their 
sum  is  90°  (§  8),  and  supplementary  when  their  sum  is  180°. 


DIRECTED  LINES 


26 


E.g.,       the  comple7nent  of  110°  is  90°  -  110°,  or  -  20°; 

"-80°"  90° -(-80°),  or  170°; 

"             "           "  ^  "  90°-^; 

and               "            "           "  -J."  90°- (-^),  or  90°  +  ^. 

"    supplement  "  135°  "  180°-  135°,  or  45°; 

"            "           "  235°  "  180°  -  235°,  or  -  55° ; 

"            "           "  A  "  180°-^; 

and              **            "          "  -^  "  180° -(-J),  or  180°  +  ^. 


EXERCISE   V 


3? 

7.  -225°?  -300°! 

8.  -415°?  -842°1 

9.  942°?  -1174°? 


In  which  quadrant  is  each  of  the  following  angle 

1.  5/3  right  angles  ?         4.    150°  ?  317°  ? 

2.  3f  right  angles  ?  5.   847°?  1111°? 

3.  17i  right  angles  ?         6.    -35°?  -140°? 

10.  Construct  the  angles  in  examples  5,  7,  9. 

11.  Give  two  positive  and  two  negative  angles,  each  of  which  is  coter- 
minal  with  45° ;  30°;  100°;  200°;   -10°;   -100°. 

Find  the  complement  and  the  supplement  of : 

12.  165°.  14.   295°  17' 14''.  16.    -  32°  14' 21". 

13.  228°.  15.   314°  22' 17".  17.    -  165°  28' 42". 
Find  the  smallest  positive  angle  coterminal  with : 

18.   420°.       19.   895°.        20.    -  330°.       21.    -  740°.       22.    - 1123°. 

20.  Positive  and  negative  lines.  If  two  lines  extend  in 
opposite  directions  and  one  of  them  is  regarded  as  positive, 
the  other  will  be  negative.  A  positive  or  a  negative  line  is 
called  a  directed  line,  and  is  read  in  the  direction  in  which  it 
extends  or  is  supposed  to  be  traced. 


Fig.  19 


Of  the  directed  line  AB,  A  is  called  the  origin  and  B  the  end. 


26 


PLANE  TRIGONOMETRY 


E.g.^  as  a  directed  line,  AB  extends  from  its  origin  A  towards  its  eru 
B,  and  CD  extends  from  its  origin  C  towards  its  end  D. 

If  we  call  AB  positive,  BA  or  CD  will  be  negative. 

Hence  AB=—  BA,  or  AB  ^  BA  =  0. 

The  nuinerical  measure  of  a  positive  or  a  negative  line  is  a 
positive  or  a  negative  real  number.     E.g.,  if  ^-S  is  four  units 
in  length  and  is  regarded  as  positive  in  direction,  then 
AB  =  +  4  units  and  BA  =  —  4  units. 

21.  Trigonometric  ratios  of  positive  or  negative  angles  of  any 
size.  In  each  of  the  four  figures  below,  let  A  denote  any 
angle,  positive  or  negative,  which  is  coterminal  with  the  angle 
XOP.  In  each  figure  a  curved  arrow  indicates  the  smallest 
positive  value  of  A,  and  a  dotted  arrow  the  smallest  in  size  of 
its  negative  values. 


Fig.  20 

From  any  point  in  the  terminal  side  OP,  as  P,  draw  MP 
perpendicular  to  the  initial  side  OX  or  OX  produced  through  0. 

In  each  of  the  four  figures  we  have  three  directed  lines,  OM, 
OP,  and  MP.  The  origin  of  the  directed  line  OM  or  OP  is  at 
the  vertex  of  the  angle,  and  the  origin  of  MP  is  in  the  initial 
side  of  the  angle  or  in  that  side  produced. 


TRIGONOMETRIC  RATIOS  27 

OM  is  regarded  as  positive  when  it  extends  in  the  direction 
of  the  initial  side  of  the  angle,  OX ;  and  hence  it  is  7iegative 
when  it  extends  in  the  opposite  direction,  OX'.  Thus  OM  is 
positive  in  fig.  a  or  d,  and  negative  in  fig.  b  or  c. 

MP  is  regarded  as  positive  when  it  extends  upward,  or  into 
the  first  or  second  quadrant ;  hence  it  is  negative  when  it 
extends  downward,  or  into  the  third  or  fourth  quadrant.  Thus 
MP  is  positive  in  fig.  a  or  b,  and  negative  in  fig.  c  or  d. 

OP  in  every  position  extends  in  the  direction  of  the  terminal 
side  of  the  angle  and  is  regarded  as  positive. 

Observe  that  in  each  figure  P  is  a  point  in  the  terminal 
side ;  MP  gives  the  distance  and  direction  of  P  from  the  initial 
side  OX,-  and  OM  gives  the  distance  and  direction  of  MP  from 
the  vertex  O. 

Ex.  1.  What  is  the  quality  of  MP  and  OM  respectively  when  A  is 
in  the  first  quadrant  ?  the  second  quadrant  ?  the  third  quadrant  ?  the 
fourth  quadrant  ? 

Ex.  2.  The  angle  A  is  in  one  of  which  two  quadrants  when  MP  is 
positive  ?    MP  negative  ?     OM  positive  ?     OM  negative  ? 

The  six  simple  ratios  (three  ratios  and  their  reciprocals) 
which  can  be  formed  with  the  three  directed  lines,  MP,  OM, 
and  OP,  are  called  the  trigonometric  ratios  of  the  angle  A. 

The  following  definitions  do  not  differ  from  those  in  §  1 
except  in  their  generality,  which  follows  from  the  use  of 
positive  and  negative  angles  and  lines. 

The  ratio  MP  /  OP  is  the  sine  of  A ; 

and  its  recij)rocal  OP  /  MP  is  the  cosecant  of  A. 

The  ratio  OM  /  OP  is  the  cosine  of  A ; 

and  its  reciprocal  OP  /  OM  is  the  secant  of  A. 

The  ratio  MP/OM  is  the  tangent  of  A ; 

and  its  reciprocal  OM/MP  is  the  cotangent  of  A. 


28  PLANE  TRIGONOMETRY 

If  two  angles  are  or  can  be  made  coterminal,  any  trigo- 
nometric ratio  of  the  one  is  evidently  equal  to  the  same 
trigonometric  ratio  of  the  other. 

Since  any  angle  denoted  by  n  ■  360°  +  A,  where  n  is  any 
real  integer,  can  be  made  coterminal  with  A,  it  follows  that 

Any  trigonometric  ratio  of  (n  •  360°  ■\-  A)  is  equal  to  the  same 
trigonometric  ratio  of  A. 

Example.  Find  a  positive  acute  angle  whose  trigonometric  ratios  are 
equal  to  those  of  420°  ;  760°  ;  1120°  ;   -  340°  ;   -  660°. 

Since  1120°,  or  3  •  360°  +  40°,  is  coterminal  with  40°,  any  trigono- 
metric ratio  of  1120°  is  equal  to  the  same  ratio  of  40°. 

22.  Laws  of  quality  of  the  trigonometric  ratios.  Two  recip- 
rocal trigonometric  ratios  must  evidently  have  the  same 
quality. 

Since  OP  is  always  positive,  the  reciprocal  ratios  sin  A  and 
CSC  A  have  the  same  quality  as  MP. 

Hence  sin  A  or  esc  A  is  positive  when  A  is  in  the  first  or  the 
second  quadrant,  and  negative  when  A  is  in  the  third  or  the 
fourth  quadrant. 

The  reciprocal  ratios  cos  A  and  sec  A  have  the  same  quality 
as  OM, 

Hence  cos  A  or  sec  A  is  positive  when  A  is  in  the  first  or  the 
fourth  quadrant,  and  negative  when  A  is  in  the  second  or  the 
third  quadrant. 

The  reciprocal  ratios  tan  A  and  cot  A  are  positive  or  nega- 
tive according  as  MP  and  OM  are  like  or  opposite  in  quality. 

Hence  tan  A  or  cot  A  is  positive  when  A  is  in  the  first  or 
the  third  quadrant,  and  negative  when  A  is  in  the  second  or  the 
fourth  quadrant. 

Observe  that  when  A  is  in  the  first  quadrant  all  its  trigo- 
nometric ratios  are  positive,  and  when  A  is  in  any  other 
quadrant  only  ttao  of  its  six  ratios  are  positive,  and  these  two 
are  reciprocals. 


TRIGONOMETRIC  RATIOS  29 

The  figure  below,  where  XOY  is  the  first  quadrant,  may 

help  to  fix  in  mind  the  very  important  laws  of  quality  given 

above. 

Y 


sin  and  esc  + 


tan  and  cot  + 


All  the  ratios  + 


cos  and  sec  + 


Fig.  21 

E.g.^  the  angle  500°  is  in  the  second  quadrant ;  hence  all  its  trigono- 
metric ratios  are  negative  except  its  sine  and  cosecant.  The  angle  —  300*^ 
is  in  the  first  quadrant ;  hence  all  its  ratios  are  positive. 

Example.  What  is  the  quality  of  each  trigonometric  ratio  of  103°? 
-135°?  235°?    -75°?  325°?   -325°?  660°?  1100°?. 

23.    Sin~^c,  C08~*c,  ....     If  sink  =  c,  then 
A  =  any  angle  whose  sine  is  c. 

The  customary  expression  for  any  angle  whose  sine  is  c  is 
sin~^c,  read  any  angle  whose  sine  is  c,  or  briefly,  angle  sine  c. 

Thus,  if  sin  A  =  c,  A  =  8in~^c,  and  conversely. 

A  similar  meaning  is  given  to  the  expressions,  cos~^b,  tan"^a, 
cot'^a,  sec'^h,  GSG~^k. 

E.g.,  sin-i(l/2)  denotes  any  angle  whose  sine  is  1/2  ;  hence  it  denotes 
any  angle  which  is  coterminal  with  30°  or  150°;  that  is, 

sin-i(l /2)=n-  360°  +  30°  or  n  ■  360°  +  150°, 

where  n  is  any  integer,  positive  or  negative,  including  0.  §  17 

Again,  tan-i  1  =  )i  •  300°  +  45°  or  n  •  360°  +  225°. 

Ex.  1.     lfA  =  cot-i(-  1),  what  are  the  values  of  A  ? 

Ex.  2.     If  J.  =  cos-i(-  1  /2),  what  are  the  values  of  ^  ? 

Ex.  3.  Given  A  =  sin-i(4/5),  to  construct  A  and  find  its  other  trigo- 
nometric ratios. 

Since  sin  ^  is  + ,  the  angle  A  is  in  the  first  or  the  second  quadrant. 


30 


PLANE  TRIGONOMETRY 


Draw  OX,  at  0  draw  OY  ±  OX,  and  lay  off  OD  equal  to  4  units. 
Through  D  draw  P'P  parallel  to  OX.     From  0  as  a  center  and  with 

a  radius  equal  to  6  units  describe  an 
arc  cutting  P'DP  in  some  points,  as 
P'  and  P.  Draw  OP  and  OR.  Then 
A,  or  sin- 1(4/5),  is  any  angle  which  is 
coterminal  with  XOP  or  XOP\ 

Hence  ^,  or  sin- 1  (4/5),  is  the  acute 
angle  XOP  +  n  •  360°,  or  the  obtuse 
angle  XOP'  +  n  •  360°,  where  n  is  any 
integer,  positive  or  negative,  including 
zero. 


Here 
Hence 


and 


OM  =  +S       and       OM'  =  -  3. 

sin  ^=4/5,  CSC  J.  =  5/4; 

cosJ.=±3/5,  sec^=±5/3; 

tanJ.=±4/3,  cotJ.  =  ±3/4. 


§21 


When,  as  above,  any  trigonometric  ratio  of  A  has  two  values  which 
are  written  together,  we  shall  consider  the  upper  sign  as  belonging  to  the 
trigonometric  ratio  of  the  least  positive  value  of  A.  Thus,  if  A  is  in  the 
first  or  the  second  quadrant,  we  shall  write  tan  A  =  ±  4/3  ;  while  if  A 
is  in  the  second  or  the  third  quadrant,  we  shall  write  tan^  =  ^  4/3 ; 
and  so  on. 


EXERCISE  VI 

Construct  A  and  find  its  other  five  trigonometric  ratios  when : 

1.  yl=sin-i(-2/3).  7.  ^  =  cos-i(- 3/7). 

2.  ^  =  tan-i(5/2).  8.  ^  =  cot-i(6/3). 

3.  ^=tan-i(-3).  9.  ^  =  cos-i(- 4/5). 

4.  ^  =cos-i(2/3).  10.  ^  =  sec-i2. 

5.  ^=sin-i(-7/8).  11.  ^  =  sec-i(- 3/2). 

6.  ^  =  tan-17.  12.  A  =  csc-i(-  5/3). 

13.    Express  each  of  the  trigonometric  ratios  of  A  in  terms  of  sin  A. 
If  sin  A  is  positive,  A  is  in  the  first  or  the  second  quadrant. 
In  fig.  22,  let  OP  =  1. 

Then  sin  A  is  the  measure  of  MP  or  M'P'. 


FUNDAMENTAL  RELATIONS  31 

Whence  MP  =  M'P'  =  sin  A, 


OM  =  ^l-sin^A,   OJf'  =  -Vl  -  sin2^. 


Hence  cos  ^  =  ±  Vl  —  sin^^,  sec  ^  =  ±  1  /v  1  —  sin2^  ; 

tan^  =  ±  sin  J./V1  —  sin^  J.,    cot^  =  ±  Vl  —  sin^^/sin^. 
If  sin  A  is  negative,  A  is  in  the  third  or  the  fourth  quadrant,  and 


cos ^  =  =F  Vl  -  sin2^,  sec  ^  =  T  1  /Vl  -  sin2^. 

24.    Fundamental  relations  between  the  trigonometric  ratios  of 
any  angle  A. 

From  the  definitions  of  the  trigonometric  ratios  of  ^ ,  we  have 

sin  A  CSC  A  =  1,  cos  A  sec  A  =  1,  tan  A  cot  A  =  1.        [1] 

MP      MP /OP      sin  A 

tan  A  = = '- = r21 

OM       OM/OP      cos  A  L  J 

Taking  the  reciprocals  of  the  members  of  r2],  we  obtain 

cot  A  =  cos  A /sin  A.  [3] 

In  each  of  the  figures  in  §  21,  we  have 

ITp'-h  Oii7^=  OP^  (1) 

Dividing  the  numbers  of  (1)  by  OP".,  we  obtain 

{MP/opy  +  {OM/  opy  =  1.  (2) 

If,  for  brevity,  we  write  (sin  .4)^  and  (cos  Ay  in  the  form 
sin^^  and  cos^^,  from  (2)  we  obtain 

sin'^A^- cos2A=  1.  [4] 

Dividing  the  members  of  (1)  by  OM  ,  we  obtain 
(MP/OMy  -\-l  =  (OP/OMy, 
or  tan'' A  +  1  =  sec'' A.  [5] 

Dividing  the  members  of  (I)  by  MP^,  we  obtain 

1  +(0M/Mpy  =  {op/Mpy, 

or  cot*A-f  1  =csc2A.  [6] 


32  PLANE  TRIGONOMETRY 

The  identities  [1]  •  •  •  [6]  express  the  more  important  of  the 
numberless  relations  that  exist  between  the  trigonometric  ratios 
of  any  angle  A. 

For  brevity  (sin  Ay,  (cos  Ay,  etc.,  are  written  in  the  form 
sin"^,  cos". 4,  etc.,  as  above,  except  when  ?i  =—  1. 

By  §  23,  sin~^c  is  used  to  denote  any  angle  whose  sine  is  c; 
hence  the  reciprocal  of  sin  A  should  never  be  written  in  the 
form  sin~M,  but  in  the  form  (sin  A)~'^  or  1/sin^. 


> 


Ex.  1.     State  identities  [!]•••  [6]  in  words. 

Ex.  2.     sec  J.  =  -  4 ;  find  the  values  of  the  other  ratios  of  A. 

Since  sec  ^  is  — ,  ^  is  in  the  second  or  the  third  qnndrant. 

§22 

sec  ^=-4;        .-.  cos  J.  =- 1/4. 

Dy[i] 

sin  ^  =  ±  Vl  -  cos2  J. 

by  [4] 

=  ±Vl-l/16  =  ±V15/4. 

.-.  CSC  ^  =  ±  4/V15  =  ±  4  V15/15. 

by[i] 

tan  A  =  sin  A /cos  A  =^  V15. 

by  [2] 

cot^=Tl/V15=  T  V15/15. 

ty[i] 

Check.  Construct  A  from  sec  A  =  —  4c,  and  then  find  the  other  trigo- 
nometric ratios  of  J.,  as  in  §  23. 

Ex.  3.   Express  the  other  trigonometric  ratios  of  A  in  terms  of  sin^. 

CSC  A  =  1  /sin  A.  by  [1] 

cos  ^  =  ±  Vl  -sin2^.  by  [4] 

.-.  sec  J.  =  ±  1/  Vl  -  sin2  ^.  by  [1] 

tan  ^  =  sin  yl  /cos  A  by  [2] 
=  ±smA/-Vl-  sin^A. 

.-.  cotA=±  Vl  -  sin2  A  /sin  A.  by  [1] 

When  sin  A  is  positive  A  is  in  the  first  or  the  second  quadrant ;  when 
A  is  in  the  first  quadrant  all  the  trigonometric  ratios  of  A  are  +  ;  when 
A  is  in  the  second  quadrant  only  sin  A  and  esc  A  are  + .  The  signs  as 
written  above  are  for  sin  A  positive. 

Check.    Find  these  relations  as  in  example  13  of  Exercise  VI.  ^     _  li*^ 


\, 


.  v^tv- '  \ 


PROOFS  OF  IDENTITIES  33 

EXERCISE  vn 

By  §  24,  compute  the  other  trigonometric  ratios  of  A,  having  given  : 

1.  smA=-2/S.  5.   tan4=-4/3.  9.    csc^=-V3. 

2.  cos^  =  1/3.  6.   cot  ^  =  -  2.  10.   sec  A  =  i. 

3.  sin  A  =  0.2.  7.    cot  A  =  3/2.  11.    tan  ^  =  -  V7. 

4.  cos^=-3/4.  8.   tan  ^  =  2.5.  12.   cos^=7n/c. 

Express  each  of  the  trigonometric  ratios  of  A  in  terms  of  : 

13.   cos^.        14.   tan^.        15.    cot -4.        16.    sec  ^.        17.   esc -4. 

25.  Proofs  of  identities.  Of  the  different  ways  of  proving 
an  identity,  the  three  following  are  the  more  common  and 
important. 

(i)  Derive   the  required  identity  from  one  or  more  known 


Ex.  1.   Prove  that  ±  Vsec2  A  +  csc^  A  =  tan  ^  +  cot  A.  (1) " 

Adding  identities  [5]  and  [6]  in  §  24,  we  obtain 

sec2  A  +  csc2  A  =  tan2  ^  +  2  +  cot2  A 

=  (tan  A  +  cot  ^)2.  by  Algebra,  [1] 

Extracting  the  square  root  of  both  members  of  the  last  identity,  we 
obtain  identity  (1). 

(ii)  Reduce  one  member  of  the  required  identity  to  the  form 
of  the  other  member,  iising  any  known  identities. 


^ 


Ex.  2.   Prove  that  a.  /- =  esc  A  —  cot  A. 

cos  A 


A  .A  1  COS -4  ,     ^,^    ^„, 

csc^-cot^  =  ^ — -  - -; — -  l>y  [1],  [3] 

sm  J.      sm  J. 

1    -    cos  ^  t,  A1  1,  TA^ 

by  Algebra,  [4] 


Vl  -cos2^ 

+  cos -4 


V(l  -  cos  AY         /l  -  cos  ^       ^     , ,    ^ 
\ ITT  —  \/  T-, 7  •     ^y  Algebra 
1  -  cos2  A        \  1  +  cos  4 


34 


PLANE  TRIGONOMETRY 


(iii)  Reduce  one  member  to  its  simplest  form,,  and  then  reduce 
the  other  member  to  the  same  form. 

When  an  identity  contains  any  other  trigonometric  ratios 
than  the  sine  and  cosine,  it  is  usually  best  in  this  method  to 
replace  these  other  ratios  by  their  values  in  terms  of  the  sine 
and  cosine. 


Ex.  3.    Prove  that 

siii2  A  tan  A  +  cos^  ^  cot  ^  +  2  sin  ^  cos  ^  =  tan  A 


cot^. 


(1) 


,  .  „  ^  sm  A  „   .  cos  A      ^   .      .  . 

First  member  =  sm^  A 1-  cos^  A h  2  sm  ^  cos  A 

cos  A  sin  A 

_  sin*  A  +  cos*  A  +  2  sin^  A  cos^  A 

~  sin  A  cos  A 

^  (sin2  A  +  cos2  A)^ 

~"       sin  A  cos  A 

=  l/(sin  A  cos^). 
Similarly  show  by  [3],  [4],  and  Algebra,  that 

second  member  =  1  /  (sin  A  cos  A). 
From  the  last  two  identities  we  obtain  identity  (1). 


by  Algebra 

by  Algebra 

by  [4] 


EXERCISE  Vra 

Prove  each  of  the  following  identities : 

1.  cos  A  tan  A  =  sin  A. 

2.  sin  ^  sec  .4  =  tan  A. 

3.  cos  J.  CSC -4  =  cot  ^. 

4.  sin  ^  cot  J.  =  cos  ^. 

5.  cos'-^^  —  sin^A  =  1  —  2  sin^^. 

6.  GOS^A  -  sin2  J.  =  2  cos^  A  -  1, 

sin  A  1  —  cos  A 


1  4-  cos  A         sin  A 
1  +  sin  J.         cos  A 


Obtain  from  [2] 
From  [2]  by  [1] 
From  [3]  by  [1] 

From  [4]  by  Algebra 
From  [4]  by  Algebra 


cos -4 


sin  A 


IDENTITIES 


36 


sec  ^  +  1 


tan -4 


From  [6]  by  Algebra 


10. 
11. 
12. 
13. 
14. 
15. 

16. 

17. 

18. 
19. 
20. 

21. 

22. 

23. 
24. 

25. 

26. 

27. 

28. 


From  [6]  by  [1] 
From  [2]  by  [1],  [5] 


tan  A         sec  A  —  \ 
sec  ^  +  tan  A  =  \/  (sec  A  -  tan  A). 
(1  +  tan2^)cos2^  =  l. 
(1  +  cot2^)sin2^  =  l. 
sin2  A  +  sin2  ^  tan2  A  =  tan2  A . 
{csc^A  —  l)csc2J.  =cos2^. 

cos*  J.  —  sin*^  +  1=2  cos2^. 

cos*^  -  sin*^  -f  1  =(cos2^  +  sin2^)(cos2^  -  sin2^)  +  1 

=  cos2^  +  (1  -  sin2^)  =  2  cos2^. 
tan2^/(l  +  tan2^)  =  sin2yl. 
cos  J. 


Vl  -  sin2^  ^ 

sin  A       ~  Vl  -  cos2^ 

C0t2^  —  C0S2^  =C0t2^  C0S2^. 

sec2^  +  csc2^  =8ec2^  csc2^. 

tan  ^  4-  cot  ^  =  sec  A  esc  A. 

cot  A  cos  A    _  cot  A  —  cos  A 
cot  A  +  cos  A  "    cot  A  cos  A 

sec2  J.  +  csc2^ 


tan  A  +  cot  A 


sec  A  CSC  A 


1/Vsec2^  -  1  =  Vcsc2^  -  1. 
1  +  tan2^      .sin2^  esc  A 


1  +  cot2  A      cos2^     cot  ^  +  tan  J. 


=  cos  A. 


cos^ 


+ 


sin  A 


sin  A  +  cos  J^. 


1  —  tan  A      1  —  cot  ^ 

sin^A  cos  A  +  cos^^  sin  ^  =  sin  2I  cos  A. 

sin2  J.  cos2^  +  cos*^  =  1  -  sin2^. 


4 


I  —  sin  ^ 
1  +  sin  4 


sec^  —  tan^. 


36 


PLANE  TRIGONOMETRY 


29.  ^^^A_  +  l±^^^^2csc^. 
1  +  cos  A  sin  A 

30.  l/(cot -4  +  taii-4)  =  sin^  cos  J.. 

31.  l/(sec^  —  tan^)  =  sec  J.  +  tan-4. 
1  —  tan  A  _  cot  A  —  \ 

32.     rzr • 

1  +  tan  A     cot  ^  +  1 

33.  ^— -^^^^=cos2^-sin2^. 
1  +  tan2  J. 

34.  CSC  A  /  (cot  A  +  tan  A)  =  cos  A. 

35.  csc^tI  (1  -  cos*^)  -  2  cot2^  =  1. 

26.  Changes  of  the  trigonometric  ratios  of  A  as  A  increases  from 
0°  to  360°.  To  simplify  this  discussion,  let  OP  have  the  same 
length  in  all  of  its  positions. 

Changes  of  sin  A.  Let  A  =  XOP,  and  let  OP  revolve  coun- 
ter-clockwise about  0  from  the  position  OX. 


Y      p 


x^: 


CHANGES  OF  TRIGONOMETRIC  RATIOS  37 

While  A  increases  from  0°  to  90°, 

MP  increases  from  0  to  OP ; 
hence  MP  /  OP,  or  sin  .1,  increases  from  0  to  +  1. 
While  A  increases  from  90°  to  180°, 

MP  decreases  from  OP  to  0 ; 
hence  MP /OP,  or  sin  A,  decreases  from  4- 1  to  0. 
While  A  increases  from  180°  to  270°, 

MP  decreases  from  0  to  —  OP ; 
hence  MP /OP,  or  sin  A,  decreases  from  0  to  —  1. 
While  A  increases  from  270°  to  360°, 

MP  increases  from  —  OP  to  0 ; 
hence  MP /  OP,  or  sin  A,  increases  from  —  1  to  0. 

Changes  of  cos  A . 

While  A  increases  from  0°  to  90°, 

OM  decreases  from  OP  to  0 ; 

hence  OM /  OP,  or  cos  A,  decreases  from  -|-  1  to  0. 

While  A  increases  from  90°  to  180°, 

OM  decreases  from  0  to  —  OP ; 

hence  OM /  OP,  or  cos  A,  decreases  from  0  to  —  1. 

Similarly  the  pupil  should  obtain  the  other  changes  of  cos  A 
given  in  the  table  below. 

Changes  of  tan  A.  Let  XOP  approach  90°  or  270°  (either 
from  a  less  or  a  greater  value)  so  that  OM  decreases  in 
size  to  1/2  its  value  the  first  second  of  time,  to  1/2  its 
remaining  value  the  next  second,  to  1/2  its  second  remaining 
value  the  third  second,  and  so  on  indefinitely.  Then,  since 
MP  increases  slightly,  MP /  OM,  or  tan  A,  more  than  doubles 
its  value  the  first  second  of  time,  more  than  doubles  its  nev?" 
value  the  next  second,  more  than  double's  its  last  value  the 
third  second,  and  so  on  indefinitely.  Thus  tan  A  will  exceed 
in  arithtnetiG  (or  absolute)  value  any  assignable  constant  num- 
ber however  great ;  that  is,  when  A  approaches  very  near  90° 
or  270°,  tan  ^  =  -h  00  or  —  00. 


38 


PLANE  TRIGONOMETRY 


Also,  when  A  =  0°,  1S0°,  or  360°,  MP  =  0,  and  therefore 
tan  ^  =  0. 

Hence  we  have  the  changes  of  tan  A  found  in  the  table  below. 

Changes  of  cot  A.  Let  XOP  approach  0°,  180°,  or  360° 
(either  from  a  less  or  greater  value)  so  that  MP  decreases 
in  size  to  1/2  its  value  the  first  second  of  time,  to  1/2  its 
new  value  the  next  second,  and  so  on ;  then  OM/MP,  or 
cot  A ,  becomes  -{-  cc  ot  —  oo  ;  that  is,  when  A  approaches  near 
0°,  180°,  or  360°,  cot  ^  =  -f  oo  or  -  oo. 

Also,  when  A  =  90°  or  270°,  OM  =  0,  and  therefore  cot  ^  =  0. 

Hence  we  have  the  changes  of  cot  A  given  in  the  table  below. 

Similarly  the  changes  of  sec  A  and  esc  A,  which  are  tabu- 
lated below,  should  be  proved  by  the  student. 

By  remembering  that  two  reciprocal  numbers  are  like  in 
quality,  that  when  the  one  increases  the  other  decreases,  and 
that  their  corresponding  values  are  reciprocals  of  each  other, 
the  changes  of  esc  A,  sec  A,  and  cot  A  are  known  from  the 
changes  of  sin  A,  cos  A,  and  tan  A  respectively. 


A  increases  from 

0°  to  90° 

90°  to  180° 

180°  to  270° 

270°  to  360° 

sin  A  varies  from 

0  to  +1 

+  1  toO 

Oto  -1 

-ItoO 

csc^      "        " 

+  00  to  +1 

+  1  to   +  00 

—  00   to    —  1 

-  1  to  -  X 

cos  A      " 

+  1  toO 

Oto  -1 

-ItoO 

Oto  +  1 

sec  ^      "       " 

+  1  to  +00 

—  00  to  —  1 

—  1  to  —  oo 

+  00  to  +  1 

tan  A  increases  from 

0  to +00 

-00  to  0 

0  to  +  00 

-  00  to  0 

cot  ^  decreases    " 

+  00  to  0 

1 

0  to  -00 

+  00  toO 

Oto  -00 

From  what  precedes  it  follows  that : 

The  tangent  or  the  cotangent  can  have  any  real  value. 

The  sine  or  the  cosine  can  have  any  value  from  —  1  ^o  -|-  1 
inclusive. 

The  secant  or  the  cosecant  can  have  any  value  from  —  cc  to 
—  1  or  from  -\-  1  to  -{-  cc  iiiclusive. 


RATIOS  OF  0°,  90°,  180°,  270^ 


39 


Observe  that  neither  the  sine  nor  the  cosine  can  have  a  value  greater 
than  +  1  or  less  than  —  1 ;  and  that  neither  the  secant  nor  the  cosecant 
can  have  any  value  between  —  1  and  +  1. 

E.g.,  +  3"/ 4  is  the  sine  of  some  angle,  the  cosine  of  some  angle,  the 
tangent  of  some  angle,  or  the  cotangent  of  some  angle ;  but  it  can  be 
neither  tlie  secant  nor  the  cosecant  of  any  angle.  Again,  —  3/2  can  be 
neither  the  sine  nor  the  cosine  of  any  angle. 

27.  Trigonometric  ratios  of  0°,  90°,  180°,  270°.  When  A  =  90° 
or  270°,  MP  =  -^  OP  or  —  OP  and  OM  =  0  ;  hence  tan  A  or 
sec  A  assumes  the  form  ±  OP/0.  Now  the  division  of  OP  by 
zero  is  impossible  ;  hence,  strictly  speaking,  90°  or  270°  has  no 
tangent  or  secant.  But  when  A  approaches  very  near  to  90° 
or  270°,  by  §  26  tan  A  or  sec  ^  is  +  x  or  —  oo  ;  hence  it  is 
customary  to  say  that  the  tangent  or  secant  of  90°  or  270°  is  oo, 
meaning  thereby  that  however  near  A  approaches  to  90°  or 
270°,  tan  A  or  sec  ^  is  +  oo  or  —  oo. 

Again,  when  .1  =  0°  or  180°,  MP  =  0  and  OM  =  +  OP  or 
—  OP;  hence  cot  ^  or  esc  J  assumes  the  form  ±  OP/0. 
Therefore,  strictly  speaking,  0°  or  180°  has  no  cotangent  or 
cosecant.  But  when  A  approaches  very  near  to  0°  or  180°, 
by  §  26  cot  A  or  esc  .1  is  -h  oo  or  —  oo  ;  hence  it  is  customary 
to  say  that  the  cotangent  or  cosecant  of  0°  or  180°  is  oo. 

The  trigonometric  ratios  of  0°,  90°,  180°,  and  270°  are  tabu- 
lated below.  To  aid  the  memory,  the  reciprocal  ratios  are 
grouped  together. 


Angle 


0° 


I      90° 


180° 


270*= 


sme  .  . 
cosecant 
cosine  . 
secant  . 
tangent 
cotangent 


0 

oo 

+  1 
+  1 

0 

oo 


+ 1 
+  1 

0 

oo 

QO 

0 


0 

QO 

- 1 
-1 

0 

00 


-1 
-1 

0 

QO 

00 

0 


40 


PLANE  TRIGONOMETRY 


Note,  Putting  OP  =  a,  tan  90°  assumes  the  form  a/0,  where  a  ji^O. 
The  form  a/0  could  be  used  as  the  tangent  of  90°;  then,  whether  we 
regarded  a/0  as  a  symbol  without  numerical  meaning,  as  a  symbol  of 
impossibility,  or  as  a  symbol  of  absolute  infinity,  when  a/0 'appeared  as 
the  tangent  of  A,  the  value  of  A  would  be  known  as  definitely  as  when 
the  tangent  of  A  is  any  finite  number. 

28.    The  trigonometric  ratios  of  —  A  in  terms  of  the  ratios  of  A. 


^X 


Fig.  24 

In  each  figure  let  A  denote  any  angle,  positive  or  negative, 
which  is  coterminal  with  XOP ;  then  —  A  will  be  coterminal 
with  XOP'.  Angle  A  is  in  the  first  quadrant  in  fig.  a,  in  the 
second  quadrant  in  fig.  b ;  and  so  on. 

Take  OP  =  OP',  and  draw  PM  ±  OX  and  P'M'  ±  OX. 

Then  in  each  figure  the  acute  angles  MOP  and  Af'OP'  will 
be  equal  in  size.  Hence  any  two  corresponding  sides  of  the 
A  OMP  and  OM'P'  will  be  equal  in  length.  Therefore,  as 
directed  lines, 

M'P'/  OP'  =  -MP  I  OP,  i.e.  sin  ( -  A)  =  -  sin  A,        (1) 
and      OM'/  OP'  =  OM/  OP,       i.e.  cos  ( -  A)  =  cos  A.  (2) 


TRIGONOMETRIC  RATIOS  OF   -  ^  41 

Dividing  (1)  by  (2),  tan(-  A)  =  -tSinA. 
Dividing  (2)  by  (1),  cot  (-  J)  =  -  cot  A. 
From  (2)  by  [1],        sec  (—A)=  sec  A. 
From  (1)  by  [1],        esc  (—  A)  =  —  esc  A. 

Identity  (1)  states  that  sin(— ^)  and  sin  A  are  arithmetic- 
ally equal  but  opposite  in  quality;  that  is,  when  sin(— ^) 
is  — ,  sin  ^  is  +  ;   and  when  sin  (—  .4)  is  -f,  sin  A  is  — . 

Identity  (2)  states  that  cos  (—A)  and  cos  A  are  arithmetically 
equal  and  like  in  quality. 

The  six  identities  just  proved  can  be  summed  up  as  follows : 

Any  trigonometric  ratio  of  —  A  is  equal  arithmetically  to  the 
same  ratio  of  A ;  but  only  the  cosines  (or  the  secants)  of  ^  A 
and  A  are  like  in  quality. 

E.g. ,       sin  ( -  35°)  =  -  sin  35°,         cos  ( -  98°)  =  cos  98°, 

tan  (-  212°)  =  -  tan  212°,     esc  (-  317°)  =  -  esc  317° 

Ex.  1.  Express  each  trigonometric  ratio  of  —  22°  in  terms  of  a  ratio 
of  22°. 

Ex.  2.  Express  each  trigonometric  ratio  of  320°  in  terms  of  a  ratio  of 
a  positive  angle  less  than  45°. 

An  angle  of  320°  is  coterminal  with  one  of  —  40°  ;  hence  any  trigono- 
metric ratio  of  320°  is  equal  to  the  same  ratio  of  —  40°  (§  21). 

Whence  sin  320°  =  sin  (-  40°)  =  -  sin  40°, 

cos  320°  =  cos  (-  40°)  =  cos  40°, 
tan  320°  =  tan  ( -  40°)  =  -  tan  40°,  etc. 

Similarly  the  trigonometric  ratios  of  any  angle  in  the  fourth  quadrant 
can  be  found  in  terms  of  those  of  some  positive  acute  angle. 

Ex.  3.  Express  each  trigonometric  ratio  of  —  325°  in  terms  of  a  ratio 
of  a  positive  angle  less  than  45°. 

An  angle  of  —  325°  is  coterminal  with  one  of  35° ;  hence 
sin  ( -  325°)  =  sin  35°,     cos  ( -  325°)  =  cos  35°,  etc. 

Similarly  the  trigonometric  ratios  of  anij  angle  in  the  first  quadrant 
can  be  found  in  terms  of  those  of  some  positive  acute  angle. 

29.  The  trigonometric  ratios  of  90°  +  A  in  terms  of  the  ratios 
of  A.      In  each  figure  let  A  denote  any  angle,  positive  or 


42 


PLANE  TRIGONOMETRY 


Degative,  which  is  coterminal  with  XOP,  and  let  POP'  =  90' 
then  A  +  90°,  or  90°  +  A,  is  coterminal  with  XOP'. 


Fig.  25 


Take  OP  =  OP',  and  draw  PM  _L  OX  and  P'J/'  _L  OX.  Then 
the  acute  angles  MOP  and  M'P'O  will  be  equal  in  size.  Hence 
any  two  corresponding  sides  of  the  A  MOP  and  M'OP'  will  be 
equal  in  length.     Therefore,  as  directed  lines, 

M'P'/OP'  =  OM/  OP,  i.e.  sin  (90°  +  A)  =  cos  A ;       (1) 

and     OM'/OP'  =  -  MP /OP,  i.e.  cos  (90°  +  A)  =  -  sin  A.  (2) 
.-.  tan  (90° +  ^4)  =  -  cot  J,  cot  (90° -f-^)  =  -  tan  A, 

sec  (90°  +  ^)  =  -CSC  .4,  esc  (90°  + ^4)  =  sec  ^. 

Since  the  angle  A  is  90°  less  than  the  angle  90°  4-  A,  the 
six  identities  just  proved  can  be  summed  up  as  follows : 

Ani/  trigonometric  7^atio  of  an  angle  is  equal  arithmetically 
to  the  co-ratio  of  this  angle  less  90°,  but  only  the  sine  (or  the 
cosecant)  of  the  first  angle  has  the  same  quality  as  the  co-ratio 
of  the  seco7id  angle. 


TRIGONOMETRIC   RATIOS  OF  90°  -{-A  43 

E.g.,  since  130°  -  90°  =  40°,  we  have 

sin  130°  =      cos  40°,    tan  130°  =  -  cot  40° ; 
cos  130°  =  -  sin  40°,     cot  130°  =  -  tan  40°. 

Ex.  1.  Express  in  terms  of  a  trigonometric  ratio  of  some  positive 
angle  less  than  45°  each  trigonometric  ratio  of  126° ;  492° ;   -  220°. 

sin  126°  =  cos  36°,     cos  126°  =  -  sin  36°,  etc.  §  29 

An  angle  of  492°  is  coterminal  with  one  of  132° ;  hence 

sin  492°  =  sin  132°  =  cos  42° ;  §§  21,  29 

cos  402°  =  cos  132°  =  -  sin  42° ;  etc. 
An  angle  of  —  220°  is  coterminal  with  one  of  140° ;  hence 

sin  (-  220°)  =  sin  140°  =  cos  50°  =  sin  40 ;  §§  21,  29,  9 

cos  (-  220°)  =  cos  140°  =  -  sin  50°  =  -  cos  40° ;  etc. 

Similarly  the  trigonometric  ratios  of  any  angle  in  the  second  quadrant 
can  be  found  in  terms  of  those  of  some  positive  acute  angle  less  than  45°. 

Ex.  2.  Express  in  terms  of  a  trigonometric  ratio  of  some  positive 
angle  less  than  45°  each  trigonometric  ratio  of  —  130° ;  230°. 

sin  (-  130°)  =  -  sin  130°  =  -  cos  40° ;  §§  28,  29 

cos  ( -  130°)  =      cos  130°  =  -  sin  40° ; 
tan  ( -  130°)  =  -  tan  130°  =      cot  40°. 
Applying  §  29  twice  in  succession  and  then  §  9  once,  we  have 
sin  230°  =      cos  140°  =  -  sin  50°  =  -  cos  40° ; 
cos  230°  =  -  sin  140°  =  -  cos  50°  =  -  sin  40° ; 
tan  230°  =  -  cot  140°  =      tan  50°  =      cot  40°  ;  etc. 
Similarly  we  can  find,  in  terms  of  the  trigonometric  ratios  of  a  positive 
angle  less  than  45°,  the  ratios  of  any  positive  or  negative  angle  in  the 
third  quadrant. 

The  principles  in  §§  9,  28,  29  have  an  important  bearing 
on  the  construction  and  use  of  trigonometric  tables  and  on 
the  solution  of  triangles.  By  them,  as  is  seen  above,  the 
trigonometric  ratios  of  any  angle  can  be  expressed  in  terms  of 
the  trigonometric  ratios  of  some  positive  angle  less  than  45°. 
Hence,  from  a  table  which  contains  the  trigonometric  ratios  of 
all  angles  between  0°  and  45°,  we  can  obtain  the  trigonometric 
ratios  of  any  angle  whatever. 


44  PLANE  TRIGONOMETRY 

30.  Trigonometric  ratios  of  complementary  and  supplementary 
angles.     Applying  §  29  twice  and  then  §  28  once,  we  obtain 

sin  (180°  -A)=  cos  (90°  -  .4)  =  -  sin  (-A)=  sin  A;  (1) 
cos  (180°  -A)=-  sin  (90°  -A)  =  -  cos  (-  ^1)  =  _  cos  A  ;  (2) 
tan  (180°-^)= -cot  (90°-^)=      tan(-^)=  -  tan^.    (3) 

Comparing  the  last  members  of  (1),  (2),  (3)  with  their  first 
members  we  have 

(i)  Any  trigonometric  ratio  of  an  angle  is  equal  arithmetic- 
ally to  the  same  ratio  of  its  supplement;  hut  only  the  sines 
(or  the  cosecants)  of  two  supplementary  angles  have  the  same 
quality. 

E.g.,  sin  150°  =     sin  30°,  tan  165°  =  -  tan  15° ; 

cos  135°  =  -  cos  45°,  cot  155°  =  -  cot  25° 

Ex.  1.  Express  in  terms  of  a  trigonometric  ratio  of  its  supplement 
each  trigonometric  ratio  of  125°  ;  143°  ;  157°. 

Comparing  the  last  members  of  (1),  (2),  (3)  with  their 
second  members  we  have  §  9  generalized ;  that  is, 

(ii)  Any  trigonometric  ratio  of  an  angle  is  equal  to  the 
co-ratio  of  its  complement. 

Ex.  2.  Applying  §  29  three  times  in  succession  and  §  28  once,  we  have 
sin  (270°±^)=  cos(180°±^)=-sin  (90°±^)  =-cos(±^)=-cos^  ; 
cos  (270°±^)  =-sin  {1S0°±A)  =-cos  (90°±^)  =  sin  (±^j  =±sin  A  ; 
tan(270°±^)=-cot(180°±^)=    tan(90°±^)=-cot  (±^)==Fcot  ^. 

Ex.  3.     Prove  (ii)  by  putting  -  ^  for  ^  in  (1)  and  (2)  of  §  29. 

31.  Trigonometric  ratios  of  n .  90  ±  A,  in  terms  of  the  ratios 
of  A.  To  obtain  in  terms  of  a  trigonometric  ratio  of  A  any 
trigonometric  ratio  of  n  •  90°  -f-  A  (where  n  is  sl  positive  inte- 
ger), we  apply  §  29  t^  times  in  succession ;  and  to  obtain  in 
terms  of  a  ratio  of  A  any  ratio  of  ^  -90°  —  ^4,  we  first  apply 
§  29  w  times  and  then  §  28  once.  In  each  case  we  change 
from  ratio  to  co-ratio  n  times  ;  hence 


TRIGONOMETRIC  RATIOS  OF  n  •  90°  ±  ^  45 

(i)    When  n  is  even,  any  trigonometric  ratio  of  n  .  90°  ±  A  is 
equal  arithmetically  to  the  same  ratio  of  A. 

(ii)    When  n  is  odd,  any  trigonometric  ratio  of  n  •  90°  ^  A  is 
equal  arithmetically  to  the  co-ratio  of  A. 

When  ^  is  a  positive  acute  angle,  any  trigonometric  ratio 
of  A  is  positive ;  hence 

(iii)  The  two  trigonometric  ratios  in  (i)  or  (ii)  will  be  oppo- 
site in  quality  when^  and  only  when,  the  ratio  o/  n .  90  ±  A  is 
negative  for  A  positive  and  acute. 

Any  positive  angle  can  be  written  in  the  form  n  •  90°  ±  A 
where  A  has  some  positive  value  less  than  45°. 

E.g.,  580°  =  6  •  90°  -f-  40" ;  here  n  is  even,  and  the  angle  is  in  the  third 
quadrant.     Hence,  by  (i)  and  §  22,  we  have 

sin  580°  =  sin  (6  •  90°  +  40°)  =  -  sin  40° ; 
cos  580°  =  cos  (6  •  90°  +  40°)  =  -  cos  40° ; 
tan  580°  =  tan  (6  •  90°  +  40°)  =  tan  40° ;  etc. 

Again,  270°  +  ^  ^3  •  90°  +  J. ;  here  n  is  odd,  and  270°  +  ^  is  in  the 
fourth  quadrant  when  A  <90°.     Hence,  by  (ii)  and  (iii),  we  have 

sin  (270°  -\-A)  =  -  cos  A,        tan  (270°  +  ^)  =  -  cot  ^  ; 
cos  (270°  ^-  A)  =  sin  A,  cot  (270°  +  ^)  =  -  tan  ^  ;  etc. 

Example.  Express  in  terms  of  a  trigonometric  ratio  of  A  each  trigo- 
nometric ratio  of  180°  +  A  ;  180°  -  A  ;  270°  -  A  ;  360°  ±  A. 

EXERCISE  IX 

Express  each  of  the  following  trigonometric  ratios  in  terms  of  the  ratio 
of  some  positive  acute  angle  less  than  45°. 

1.  sin  168°.  6.  cos (-84°).  11.  cot  1054°. 

2.  tan  137°.  7.  tan  (-246°).  12.  sec  1327°. 

3.  cos  287°.  8.  cos  (-428°).  13.  esc  756°. 

4.  sin  834°.  9.  cos  1410°.  14.  tan  (- 196°  54'). 

5.  sin  (-65°).  10.  tan  1145°.  15.  cot  (- 236°  21'). 


46  PLANE  TRIGONOMETRY 

16.  Prove  sin  420°  •  cos  390°  +  cos  ( -  300°)  •  sin  ( -  330°)  =  1. 

17.  Prove  cos  570°  •  sin  510°  -  sin  330°  •  cos  390°  =  0. 

32.    Trigonometric  lines  representing  the  trigonometric  ratios. 

Any  trigonometric  ratio  is  a  positive  or  a  negative  number, 
but  it  can  always  be  represented  by  a  directed  line,  as  below. 
Let  A  denote  any  angle  cotenninal  with  Z.  XOP  in  each  of 
the  four  figures.  Take  OP  as  a  positive  unit  line,  and  draw 
PMl.  OX. 

Then  sin  A  =  MP  J  OP  =  the  numerical  measure  of  MP ;  (1) 
hence  sin  A  is  represented  hy  the  directed  line  MP. 

Also,  cos  A  =  OM I  OP  =  the  numerical  measure  of  OM;  (2) 
hence  cos  A  is  represented  by  the  directed  line  OM. 

T  b 


0         M      ; 


X' 


1 

y 

^ 

T 

M  ^ 

^ 

X 

/- 

0 

\ 

y^ 

N 

,y 

/ 

/• 

\^ 

c       c__^ 


Fig.  26 


TRIGONOMETRIC  LINES  47 

To  obtain  directed  lines  which  shall  represent  the  four  other 
trigonometric  ratios,  draw  OY 1.  OX  at  0,  and  take 
OX  =  OY  =  OP  =  a  positive  unit  line. 

At  X  draw  XT-JL  OX,  at  Y  draw  YC  _L  OF,  and  prolong  each 
until  it  meets  the  final  side  OP  (produced  through  P  or  O)  in 
some  point  as  T  or  C. 

According  to  the  laws  assumed  in  §  21  for  the  quality  of 
MP,  OM,  and  OP,  the  directed  line  AT  is  positive  or  negative 
according  as  it  extends  upward  or  downward  from  its  origin 
A;  YC  is  positive  or  negative  according  as  it  extends  to  the 
right  or  to  the  left  from  its  origin  F;  and  OT  or  OC  is  posi- 
tive or  negative  according  as  it  extends  from  the  origin  0  in 
the  direction  of  the  final  side  OP  or  in  the  opposite  direction. 

E.g.,  XT  or  YC  is  +  in  fig.  a  or  c,  and  —  in  fig.  h  or  d. 
or  is  +  in  fig.  a  or  tZ,  and  —  in  fig.  6  or  c. 
OC  is  +  in  fig.  a  or  6,  and  —  in  fig.  c  or  d. 

In  each  figure  the  triangles  OMP,  OXT,  and  OYC,  being 
mutually  equiangular,  are  similar. 

In  each  of  the  four  figures  we  find  that 
tan^  =  MPjOM—  XT /OX  =  the  numerical  measure  of  XT ;  (3) 
hence  tan  A  is  represented  by  the  directed  line  XT. 
cot  A  =  OM/MP  =  YC/OY=the  numerical  measure  ofYC;  (4) 
hence  cot  A  is  represented  by  the  directed  line  YC. 
sec  A  =  OP /  OM  =  OT/OX  =  the  numerical  measure  of  OT ;  (5) 
hence  sec  A  is  represented  by  the  directed  line  OT. 
CSC  A  =  OP  I  MP  =z  OC  /  OY  =  the  numerical  measure  of  OC  ;  (6) 
hence  esc  A  is  represented  by  the  directed  line  OC. 

The  directed  lines  which  represent  the  trigonometric  ratios 
of  an  angle  are  called  the  trigonometric  lines  of  that  angle. 

The  relations  in  (1)  to  (6)  can  be  written  briefly 
sin  A  =  MP,  cos  A  =  OM,  tan  A  =  XT, 
cot  A  =  YC,   sec  A  =  OT,    esc  A  =  OC, 


48  PLANE  TRIGONOMETRY 

Since  the  trigonometric  lines  represent  graphically  the  trigo- 
nometric ratios,  or,  in  other  words,  the  trigonometric  ratios  are 
the  numbers  which  measure  the  trigonometric  lines,  it  follows 
that  if  we  prove  any  relation  between  the  trigonometric  lines, 
we  know  that  the  same  relation  exists  between  the  correspond- 
ing trigonometric  ratios,  and  vice  versa. 

33.    Use  of  trigonometric  lines  in  proofs  and  discussions.     To 

fix  in  the  pupil's  mind  the  trigonometric  lines  which  represent 
the  trigonometric  ratios,  to  help  familiarize  him  with  the  use 
of  directed  lines  to  represent  positive  and  negative  real  num- 
bers, and  to  show  him  how  the  use  of  the  trigonometric  lines 
sometimes  simplifies  trigonometric  proofs  and  discussions,  we 
give  below  illustrative  examples,  which  can  be  taken  or  omitted 
at  the  option  of  the  teacher. 

Ex.  1.     Using  trigonometric  lines,  prove  the  relations  in  §  24. 
In  each  of  the  four  figures  in  §  32  we  have 

JWP VOJW"^  =  0P^        .-.  sin2^  +  cos2^=  1; 
■qx^  +  ^T^  =  ~OT^,  .-.  1  + tan2^  =  sec2^; 

'6Y^+YC'^  =  ~0C'^,  .-.  1  4-COt2  ^=CSC2^; 

XT/ OX  =  MP/ OM,  .'.  tan  A=8mA/cosA; 

YC/  OY  =  OM/MP,  .-.  cot  ^  =  cos  ^  /sin  A  ; 

OT/OX='OP/OM,  .-.  sec  J.  =  l/cos^; 

OC/OY=  OP/ON,  .'.  CSC  A  =  l/siuA. 

Ex.  2.  Using  the  trigonometric  lines,  determine  the  quality  of  each 
trigonometric  ratio  in  each  quadrant. 

In  the  figures  of  §  32,  the  quality  of  MP,  or  sin  A,  is  easily  deter- 
mined.    So  also  is  the  quality  of  OM,  or  cos -4. 

XT,  or  tan  A,  is  positive  when  A  is  in  the  first  or  the  third  quadrant, 
and  negative  when  A  is  in  the  second  or  the  fourth  quadrant. 

or,  or  sec  A,  extends  in  the  direction  of  OP,  and  is  therefore  positive 
when  A  is  in  the  first  or  the  fourth  quadrant ;  and  OT,  or  sec  A,  extends 
in  the  direction  opposite  to  that  of  OP,  and  is  therefore  negative  when  A 
is  in  the  second  or  the  third  quadrant. 

In  like  manner  determine  the  quality  of  YC,  or  cot  A,  and  of  OC,  or 
esc  A. 


TRIGONOMETRIC  LINES 


49 


Ex.  3.  Using  trigonometric  lines,  trace  the  changes  of  the  trigono- 
metric ratios  of  A  while  A  increases  from  0°  to  360°. 

In  the  figures  of  §32,  the  changes  of  JfP,  or  sin -4,  and  of  OM^  or 
cos^,  are  easily  followed. 

While  A  increases  from  0°  to  90°  (fig.  a),  XT  beginning  at  zero  increases 
without  limit  as  A  approaches  90°;  i.e.  tan  A  increases  from  0  to  +  oo. 

While  A  increases  from  90°  to  180°  (fig.  6),  XT  is  at  first  of  infinite 
length  and  negative,  and  becomes  0  when  A  =  180°  ;  i.e.  tan  A  increases 
from  —  GO  to  0. 

While  A  increases  from  180°  to  270°  (fig.  c),  XT  beginning  at  zero 
increases  without  limit  as  A  approaches  270° ;  i.e.  tan  A  increases  from 
0  to  +  00. 

While  A  increases  from  270°  to  360°  (fig.  d),  XT  is  at  first  of  infinite 
length  and  negative,  and  becomes  0  when  A  =  360° ;  i.e.  tan  A  increases 
from  —  GO  to  0. 

In  like  manner  the  student  should  trace  the  changes  of  cot^,  sec  J., 
and  CSC  A. 

Ex.  4.  Using  the  trigonometric  lines,  find  the  trigonometric  ratios  of 
180°  -  A  and  180°  -f  ^  in  terms  of 

those  of  A,  when  A  is  in  the  first  C^  Y  C 

quadrant. 

Let  the  angles  XOP,  M'OP\ 
and  M"OP"  be  equal  in  size  ;  then 
if  A  is  coterminal  with  XOP^ 
180°  —  A  will  be  coterminal  with 
XOP",  and  180°  +  A  will  be  coter- 
minal with  XOP". 

Draw  the  trigonometric  lines  of 
A,  180°  -  A,  and  180°  +  A. 


Fig.  27 


Then 


Again, 


M'P'  =  MP, 
OM'  =  -  OM, 

XT'  =  -  XT, 

YC  =  -  YC, 

0T=-  OT, 

OC  =  OC, 

M"P"  =  -  MP, 

OW  =  -  OM, 
XT  =  XT, 
YC  =  YC, 


.-.  sin  (180°  -  ^)  =  sin  ^  ; 
.-.  cos  (180°  -  ^)  =  -  cos  ^ 
.-.  tan  (180°  -  ^)  =  -  tan  ^ 
.-.  cot  (180°  -  ^)  =  -  cot  ^ 
.-.  sec  (180°  -  ^)  =  -  sec  ^ 
.-.  CSC  (180°-  ^)  =  csc^. 
.-.  sin  (180°  +  ^)  =  -  sin  ^ 
.-.  cos  (180°  +  ^)  =  -  cos  4 
.-.  tan  (180°  +  ^)  =  tan  4  ; 
.-.  cot  (180°  +  A)  =  cot  A. 


50  PLANE  TRIGONOMETRY 

OT  and  OC  are  both  negative  when  the  angle  is  XOP". 

Hence  sec  (180°  ■\-  A)=  -  sec  A,     esc  (180°  -\-  A)  =  -  esc  A. 

A  similar  proof  could  he  given  when  A  is  in  any  one  of  the  other 
three  quadrants. 

Ex.  5.     Using  the  trigonometric  lines,  find  the  trigonometric  ratios  of 
—  A  in  terms  of  those  of  A. 
In  each  figure  of  §  28,  let 

OP  =  OP'  =  a  positive  unit  line. 
Then  MP  =  sin  ^,  OM  =  cos  A, 

M'P'  =  sin  (-  A),  OM'  =  cos  (-  A). 

But  in  each  figure  we  have 

M'P'  =  -  MP,  OM'  =  OM. 

Hence        sin  (— ^)  =  —  sin^,      cos  (— ^)  =  cos  J..  (1) 

From  the  identities  (1)  we  can  obtain  the  other  relations  as  in  §  28. 

The  student  should  draw  the  other  trigonometric  lines  of  A  and  —  A 
in  each  of  the  four  figures  in  §  28,  and  prove  the  last  four  identities  by 
the  use  of  these  lines. 

Ex.  6.     Using  the  trigonometric  lines,  find  the  trigonometric  ratios  of 
90° -f  J.  in  terms  of  those  of  A. 
In  each  figure  of  §  29,  let 

OP  =  OP'  =  a  positive  unit  line. 
Then  MP  =  sin  ^ ,  OM  =  cos  ^, 

M'P'  =  sin  (90°  +  A),      OM'  =  cos  (90°  +  A). 
But  in  each  figure  we  have 

M'P'  ~  OM,  OM'  =  -  MP. 

Hence  sin  (90°  + ^)=  cos ^,    cos(90°  +  ^)=  -  sin  ^.  (1) 

From  the  identities  (1)  we  can  obtain  the  other  relations  as  in  §  29. 

But  the  student  should  draw  the  other  trigonometric  lines  of  A  and 
90°  +  ^  in  each  of  the  four  figures  in  §  29,  and  prove  the  last  four  identi- 
ties by  the  use  of  these  lines. 

Ex.  7.  Using  the  trigonometric  lines,  find  the  trigonometric  ratios  of 
90°  —  ^  in  terms  of  those  of  A,  when  A  is  in  the  first  quadrant. 


TRIGONOMETRIC  LINES  61 

In  the  figures  of  §  32,  let  X  be  the  origin  of  arcs,  and  let  the 
arc  be  positive  or  negative  according  as  its  generating  point 
moves  counter-clockwise  or  clockwise.  Then  the  arc  in  each 
figure  will  have  the  same  numerical  value  in  degrees  as  the 
angle  which  it  subtends  at  the  center,  and  the  trigonometric 
lines  of  the  angle  in  each  figure  can  be  regarded  as  trigono- 
metric lines  of  the  arc,  and  the  ratios  which  these  lines  repre- 
sent can  be  regarded  as  trigonometric  ratios  of  these  arcs. 
Hence,  if  the  number  of  degrees  in  a  unit  arc  is  equal  to  the 
number  of  degrees  in  an  angle,  the  arc  and  the  angle  have  the 
same  trigonometric  lines  or  ratios. 


CHAPTER   III 
TRIGONOMETRIC  RATIOS  OF  TWO  ANGLES 

34.    Sine  and  cosine  of  the  sum  of  two  angles.     Let  XOR  and 

ROC  \)Q  any  two  positive  acute  angles. 

Then  Z  XOR  +  Z  ROC  =  Z  XOC. 

Let  A  denote  any  angle,  positive  or  negative,  coterminal 
with  Z  XOR,  and  B  any  angle  coterminal  with  Z  ROC. 
Then  the  sum  A  -\-  B  will  be  coterminal  with  Z.  XOC. 


>X 


Fig.  28 


The  sum  A  -\-  B  may  be  in  the  first  quadrant,  as  in  fig.  a,  or 
in  the  second  quadrant,  as  in  fig.  h. 

In  each  figure,  from  any  point  on  OC,  as  P,  draw  PN  ±  OR 
and  PMl.  OX ;  also  draw  NQ  ±  OX,  and  ND  _L  MP. 

Then  the  triangles  DPN  and  QON  will  be  similar. 

Now  MP=QN-\-  DP. 

By  §  21,  QN=BmAON. 

Again,  DP  /  NP  =  OQ/ ON  =  cos  A; 

whence  i>P  =  cos  ^  •  NP. 

62 


ADDITION  FORMULAS  63 

Hence  MP  =  sin  ^  •  ON  +  cos  ^  •  NP. 

.'.  MP  I  OP  =  sin  ^  •  ON  I  OP  +  cos  ^  •  NP/OP.       (1) 
Substituting  for  the  ratios  in  (1)  their  names,  we  have 

sin  (A  +  B)  =  sin  A  cos  B  +  cos  A  sin  B.  [7] 

Again,  OM=OQ-  DN 

=  cos  A  •  ON  —  sin  A  •  NP. 
.'.  OM/ OP  =  cos  ^  .  ON/ OP  -  sin  ^  •  NP/OP. 
.'.  cos  (A  +  B)  =  cos  A  cos  B  —  sin  A  sin  B.  [8] 

Observe  that  thus  far  [7]  and  [8]  are  proved  only  when  the  angles 
A  and  B  are  both  in  the  first  quadrant.  In  §  35  it  will  be  shown  that 
these  relations  hold  true  in  whatever  quadrant  J.  or  1?  is. 

35.   General  proof  of  [7]  and  [8]. 


sin(.4  -f  90°  +  B)  =  sin  (90°  -\- A -\- B) 

=  cos(A-\-B)  §29 

=  cos  A  cos  B  -\-(—  sin  A)  sin  B  by  [8] 

=  sin  (A  +  90°)  cos  B  +  cos  (A  +  90°)  sin  B.       (1) 


Again,  cos  (A  +  90°  -\- B)  =  cos  (90°  -\- A  -^  B) 

=  -  sin  (A  i-B)  §  29 

=  (—  sin  A)  cos  B  —  cos  A  sin  B  by  [7] 

=  cos  (A  +  90°)  cos  B  -  sin  (A  -f  90°)  sin  B.       (2) 

Now  in  whatever  quadrant  ^4  is,  ^  +  90°  is  in  the  next 
quadrant.  Hence,  from  (1)  and  (2),  it  follows  that  if  [7]  and 
[8]  are  true  when  A  is  in  any  one  quadrant,  they  are  true 
also  when  A  is  in  the  next  quadrant.  But,  by  §  34,  [7]  and 
[8]  are  true  when  A  is  in  the  first  quadrant;  hence  they  are 
true  when  A  is  in  the  second  quadrant;  and  so  on.  Hence 
[7]  and  [8]  hold  true  in  whatever  quadrant  A  is. 

The  same  reasoning  applies  to  B.  Hence  [7]  and  [8]  hold 
true  for  all  values  of  A  and  B,  positive  or  negative. 


64  PLANE    TRIGONOMETRY 

Formulas  [7]  and  [8],  often  called  the  addition  formulas,  are 
very  important  and  should  be  memorized. 

So  many  theorems  can  be  deduced  from  the  formulas  [7]  and  [8]  that 
they  are  often  called  the  fundamental  formulas  of  trigonometry. 

EXERCISE  X 

1.  State  in  words  identities  [7]  and  [8],  as  generalized  in  §  35. 

The  sine  of  the  sum')  _  f  sin  first  ■  cos  second 
of  any  two  angles  j  ~~  t_  +  cos  first  •  sin  second. 

2.  sin  75°  =  sin  (30°  +  45°) 

=  sin  30°  cos  45°  +  cos  30°  sin  45°  by  [7] 

_  1   V2     V3  V2  _  V2  +  Ve 

3.  Putting  75°  =  30°  +  45°  and  using  [8],  find  cos  75°. 

4.  Putting  15°  =  45°  +  (-  30°),  find  sin  15°  and  cos  15°. 

5.  Putting  15°  =  60°  +  (-  45°),  find  sin  15°  and  cos  15°. 

6.  Putting  90°  =  60°  +  30°,  find  sin  90°  and  cos  90°. 

7.  Putting    0°  =  45°  +  (-  45°),  find  sin  0°  and  cos  0°. 

A  and  B  being  positive  acute  angles,  find  the  values  of  sin  {A  +  B) 
and  cos  {A  +  B),  having  given 

8.  sin^  =  2/5,  cosJB=:l/3.        9.    sin  ^  =  2/3,  cos  B  =  1/4. 

10.  Putting  90°  +  ^  for  ^  in  [7],  deduce  [8]. 

11.  Putting  90°  +  A  ioT  A  in  [8],  deduce  [7]. 

12.  Prove  [7]  and  [8],  using  trigonometric  lines,  A  and  B  being  in 
the  first  quadrant. 

Take  OP  =  +  1. 

Then  MP  =  sin  {A  +  B), 

ON  =  cos  B,     NP  =  sin  B. 
.-.  DP  =  NP  cos  DPN  =  sin  BcosA. 
QN  =  ON  sin  A  =  cos  5  sin  ^. 
.-.  sin  {A  -\-  B)=  QN  +  DP  =  sin  AcosB  -h  cos  A  sin  B. 


SUBTRACTION  FORMULAS  55 

36.    Sine  and  cosine  of  the  difference  of  two  angles.     Substitut- 
ing —  B  for  B  in  [7],  we  have 

sin  (.4  —  B)  =  sin  A  cos  (—  B)  +  cos  A  sin  (—  B). 
,'.  sin  (A  —  B)  =  sin  A  cos  B  —  cos  A  sin  B.  [9] 

Substituting  —  B  foT  B  in  [8],  we  have 

cos  (A  —  B)  =  cos  A  cos  (—  B)  —  sin  A  sin  (—  B). 
.'.  cos  (A  —  B)  =  cos  A  cos  B  +  sin  A  sin  B.  [10] 

Formulas   [9]   and   [10]   are   often    called   the   subtraction 
formulas. 

EXERCISE   XI 

1.  State  in  words  identities  [9]  and  [10]. 

The  sine  of  the  difference  \_(      sin  first  •  cos  second 
of  any  two  angles  J  ~  I  ~~  (^os  first  •  sin  second. 

2.  Putting  15°  =  45°  -  30°,  find  sin  15°  by  [9]  and  cos  16°  by  [10]. 

3.  Putting  15°  =  G0°  -  45°,  find  sin  15°  and  cos  15°. 

A  and  B  being  positive  acute  angles,  find  the  values  of  sin  {A  —  B)  and 
cos  {A  —  B),  having  given 

4.  sin  ^  =  1/4,  sin  5=  1/3.  5.    cos  ^  =  2/3,  cos  5  =  3/4. 

Prove  each  of  the  following  identities : 

6.  sin  {A  +  B)  sin  {A  -  B)  =  sm^  A  cos^  B  -  cos2  Asin^B 

+  (sin2 ^  sin2  B  -  sin2  A  sin2  B) 
=  sin2  A  (cos2  B  +  sin2  B)  -  sin2  5(cos2  A  +  sin^  A) 
=  sin2^  -sin^B. 

Observe  that  sin 2  J.  sin^  B  —  sin2  ^  sin 2  B  is  added  above  as  one  form 
of  zero. 

7.  cos  (A  +  B)  cos  {A  -  B)  =  cos^A  -  sin2  B. 

8.  sin  {A  +  B)  cos  B  -  cos  {A  +  B)  sin  B  =  sin  A. 

9.  sin  {A  -{-  B)  +  cos  {A  -  B)  =  (sin  A  +  cos  A)  (sin  B  +  cos  B). 
10.  sin  A  cos  {B  -  C)  -  sin  Bco8{A  +  C)  =  sin  {A  -  B)  cos  C. 


66  PLANE  TRIGONOMETRY 

11.  tan  A  +  t&nB  =  sin  {A  +  ^)/(cos  A  cos  B). 

12.  cot  JB  -  cot  ^  =  sin  {A  -  B)/{sin  A  sin  B). 

13.  Prove  [9]  and  [10]  geometrically,  using  trigonometric  lines,  when 
A,  B,  and  A  —  B  are  in  the  first  quadrant. 

Let  XOB  and  BOChe  any  two  acute  angles,  Z  BOC 
being  negative  and  Z  XOC  being  positive. 
Then  Z  XOC  =  Z  XOB  +  Z  BOC. 

Let  A  denote  any  angle  coterminal  with  XOB,  and 
—  B  any  angle  coterminal  with  BOC. 
1>X       Then  A  +  {-  B),  or  A  -  B,  will  be  coterminal 
with  XOC. 

Take  OP  equal  to  +  1. 
Draw  PM  ±  OX,  PN  ±  OB,  NQ  ±  OX,  and  PD  _L  QN. 

Now  NP  =  sin  {-  B)  =  -  sin  5,  ON  =  cos  (-  J5)  =  cos  B, 

and  sin  {A  -  B)  =  MP  =  QN  -  DN. 

Also,  QN=  ON'  sin  JTOE  =  cos  ^ sin  ^, 

and  DN  =  ( -  NP)  cos  DiVP  =  sin  BcosA. 

.:  sin  (^  -  J5)  =  QiV  —  I)N  =  sin  J.  cos  -B  -  cos  A  sin  B. 
Again,  OQ  =  OJV  cos  XOE  =  cos  BcosA, 

and  DP  =  {-  NP)  sin  DJVP  =  sin  jB  sin  ^. 

.-.  cos  {A  -  B)=:  0M=  0Q  +  DP  =  cos  ^  cos  ^  +  sin  A  sin  B. 

37.  Tangent  of  the  sum  and  difference  of  two  angles.  Divide 
the  members  of  [7]  by  those  of  [8]  ;  then  by  [2]  we  have 

.  ,        ^-        sin  A  cos  B  +  cos  A  sin  B 

tan  (A  +  B)  = : — - — -. — -•  (1) 

^  ^       cos  A  cos  B  —  sm  A  sm  B  ^  ^ 

To  express  tan  (A  +  B)  in  terms  of  tan  A  and  tan  B  we 
divide  the  numerator  and  denominator  of  the  fraction  in  (1) 
by  cos  A  cos  B ;  then  by  formula  [2]  we  obtain 

.«x        tan  A  4-  tan  B  ^^  ^  _, 

tan  (A  +  B)  = —^ -•  fill 

^     ^    ^      1  -  tan  A  tan  B  *-     -■ 

Substituting  —  B  for  B  in  [11],  we  obtain 

, .       „.        tan  A  —  tan  B 

tan  (A  -  B)  = ; -•  [12] 

^  ^      1  +  tan  A  tan  B  •-     -■ 


TRIGONOMETRIC  RATIOS  OF  2  A  67 


EXERCISE  Xn 

1.  State  in  words  identities  [11]  and  [12]. 

The  tangent  of  the  sum  \  _  f     the  sum  of  their  tangents 
of  any  two  angles      j  ~  \  i  _  product  of  their  tangents 

2.  Putting  75°  =  45°  +  30°,  find  tan  75°  by  [11]. 

3.  Putting  15°  =  60°  -  45°,  find  tan  15°  by  [12]. 

4.  If  tan  ^  =  -  1/2  and  tan  B  =  3,  find  tan  {A  +  E)  and  tan  {A  -  B). 

5.  If  tan  ^  =  -  2  and  tan  B  =  -  3,  find  tan  {A  +  B)  and  tan  {A  -  B). 
Prove  each  of  the  following  identities  : 

a    ^o^MRo,    jx_l+tan^  ,  _.  _cotAcotB  -  1 

6.  tan  (45°  +  ^)  = •  8.    cot  (A  +  B)  = 

1  -  tan  ^  cot  B  +  cot  A 

r,    4.      /AKo       A\      1  ~  t^°  ^  o        ^  / .       «v      cot  A  cot  B  4-  1 

7.  tan  (45°-^)  =  - -•  9,    cot(A-B)  = — — 

1  +  tan  ^  cot  B  -  cot  A 

10.  Prove  identity  [12]  by  dividing  [9]  by  [10]. 

11.  Prove  the  identities  in  examples  8  and  9  by  taking  the  reciprocals 
of  the  members  of  [11]  and  [12]  respectively. 

12.  Find  tan  {A  +  B)  and  tan  {A  -  B)  in  terms  of  cot  A  and  cot  B. 

13.  Find  cot  {A  +  B)  and  cot  {A  -  B)  in  terms  of  tan  A  and  tan  B. 

38.    Trigonometric  ratios  of  twice  an  angle  in  terms  of  the  ratios 

of  the  angle.     Substituting  A  for  B  in  [7],  we  have 

sin  (A  -{-  A)  =  sin  A  cos  A  -{-  cos  ^  sin  ^  ; 

tliat  is  sin  2  A  =  2  sin  A  cos  A.  [13] 

Substituting  ^  for  5  in  [8],  we  obtain 

cos  2  A  =  cos*  A  —  sin*  A        (i)  ^ 

=  1  -2  sin*  A  (ii)   ►  [14] 

=  2cos*A-l.         (iii) 

To  derive  Qi)  or  (iii)  from  (i),  we  use  identity  [4]. 

Substituting  A  for  B  in  [11],  we  obtain 

^      ^  ^  2  tan  A  _ .  ^_ 

tan2A  = — -•  [15] 

1  -  tan*  A  •-     -■ 


58  PLANE  TRIGONOMETRY 


EXERCISE  XIII 

1.  State  in  words  identities  [13],  [14],  and  [15]. 

sin  twice  an  angle  =  2  sin  angle  •  cos  angle. 
cos  twice  an  angle  =  (cos  angle)^  —  (sin  angle)^. 

2.  From  the  trigonometric  ratios  of  30°,  find  sin  60°,  cos  60°,  tan  60°. 

3.  From  the  trigonometric   ratios   of  60°,   find  sin  120°,    cos  120°, 
tan  120°. 

4.  Express  sin  6^,  cos  6^,  tan  6  J.  in  terms  of  the  trigonometric 
ratios  of  3  -4. 

5.  Express  sin  3^,  cos  3^,  tan  3^  in  terms  of  the  trigonometric 
ratios  of  S  A/2. 

Prove  each  of  the  following  identities  : 

_        ^  _    .      cot2  A  -1  _      .,^       1-COS2J. 

6.  cot  2  ud  = 8.    sm2  A  = 

2  cot  J.  2 

7.  csc2^  =  (sec  J.csc^)/2.  9.    cos^^  =(1  +  cos2  ^)/2. 
sec2  A          1  +  tan2  A 


10.   sec 2^  = 


2-sec2^       l-tan2^ 

11.  cos4^=2cos2  2^- 1  =  2(1  -2sin2^)2_i 

=  8sin4^ -8sin2^  +  1. 

12.  sin  4  J.  =  4  sin  ^  cos  ^  —  8  sin^  ^  cos  ^. 

39.  Trigonometric  ratios  of  half  an  angle  in  terms  of  the  cosine 
of  the  angle.  Solving  (ii)  and  (iii)  of  [14]  for  sin^^l  and 
cos^  A  respectively  and  putting  A/2  for  A,  we  obtain 

.A  /I  —  cos  A  _. -_ 

,                                              A          /I  +  cos  A  ^.^_ 

and  COS  -  =  ^— L- ,  [17] 

Divide  [16]  by  [17],    tan |  ^  V^^^-  [18] 


TKIGONOMETRIC  RATIOS  OF  A/2  69 


EXERCISE  XIV 

1.  State  in  words  identities  [16],  [17],  and  [18]. 

•     17^             7                        ..    r  1  —  cos  angle  _,  ^_ 

sin  Jialf  an  angle  =  square  root  of ^— .  [16] 

2.  Find  sin  22i°,  cos  22|°,  tan  22 1°,  from  cos  46°. 
,         cos 45°  _      /I  -V2/2  _  V2-V2 


in22i°  =  J- 


2  \  2  2  * 

3.  Find  sin  16°,  cos  15°,  tan  15°,  from  cos  30°. 

4.  cos  A  =  1  /S  ;  find  the  sine,  cosine,  and  tangent  of  -4/2. 

5.  cos  A  =  a;  find  the  sine,  cosine,  and  tangent  of  ^/2. 

6.  Express  sin  A,  cos  A,  and  tan  A  in  terms  of  cos  2  A. 

7.  Express  sin  2  A,  cos  2  A,  and  tan  2  ^  in  terms  of  cos  4  J.. 

8.  Express  sin  SA,  cos  3  A,  and  tan  3  ^  in  terms  of  cos  6  A. 
Prove  each  of  the  following  identities  : 

^A  _1  +  cosA_/    sin  A     \^  _  /I  +cosA\^ 
2  "1  —  cos  J.~  \1  — cosJ./   ~\    sin  J.     / 

•i/x    X     o-^      /     sin^     \2     /l-cos^\2 

10.  tan2  —  =  ( )  =  ( )  =  (esc  A  -  cot  A)^. 

,1  ^A        2  sec  J.  ^  _  ^A        2  sec  ^ 

11.  sec2— = : 12.    csc2  — = 


2      sec  -4  +  1  2      sec  A  —  \ 

13.  Express  cos*  A  in  terms  of  cos  2  A  and  cos  4  ^. 

(cos2^)2  =  (i+i  cos  2^)2 

=  i+^cos2J.  +  J  cos2  2  ^ 

=  ^  +  |cos2^  +  ^(1  +  icos4^). 

.-.  cos* -4  =  f  +  i  cos  2 -4  +  i  cos  4 .4. 

14.  Prove  sin* -4  =  f  —  |  cos  2  -4  +  |-  cos  4  J.. 

15.  Prove  sin2  A  cos2 .4  =  i  —  i  cos  4 .4. 
Suggestion,     sin^  .4  cos2  ^  =  (sin  A  cos  .4)2  =  i  sin2  2  A. 

16.  Prove  sin2 .4  cos*  A^^^  +  I  sin2  2  ^  •  cos  2  ^  —  y^^  cos  4 .4. 
Suggestion.     sin2  ^  cos*  -4  =  (sin  -.4  cos  -4)2  ■  cos2  A. 


60  PLANE   TRIGONOMETRY 

40.    Sum  and  difference  of  sines  and  cosines.      Adding  and 
subtracting  [7]  and  [9],  and  [8]  and  [10],  we  obtain 

sin  (A  +  B)-{-  sin  (A  -  B)  ~      2  sin  ^  cos  ^ ;  (1) 

sin  (A  +  B)-  sin  (^1  -  B)  =       2  cos  A  sin  B ;  (2) 

cos  (A  -\-  B)-\-  cos  (A  —  B)  =      2  cos  A  cos  B ;  (3) 

cos  (A  -{-  B)-  cos  (A  —  B)  =  —2smA  sin  B.  (4) 

Let         A  -^B=  C       and       A  -  B  =  D.  ') 

Then               A  =  (C+D)/2,           B  =  (C  -D)/2.j  ^^'^ 


Substituting  in  (1)  •  •  •  (4)  the  values  in  (5),  we  obtain 

C  4-  D        C  —  D 

sinC  +  sinD=      2  sin — ■ — cos 

2                  2 

[19] 

C  +  D        C  —  D 

sin  C  —  sin  D  =      2  cos  — - —  sin  — 

2                    2 

[20] 

C+DC-D 

cos  C  +  cos  D  =      2  cos  — - —  cos  — 

[21] 

C  +  D        C  —  D 

cos  C  —  cos  D  =  —  2  sin  — ! —  sin 

9                            9 

[22] 

By  formulas  [19]  •  •  •  [22],  a  sum  or  a  difference  of  the  sines 
or  the  cosines  of  two  angles  is  transformed  into  a  product. 
Hence  these  formulas,  often  called  product  formulas,  are  useful 
in  adapting  other  formulas  to  the  use  of  logarithms. 

E.g.,      sin7  J.  +  sin5^  =  2sini(7^  +  5^)cosi(7^  -  5A) 

=  2  sin  6  A  cos  A . 
.-.  log  (sin  7  J.  +  sin  5  J.)  =  log  2  +  log  sin  6  ^  +  log  cos  A. 
Again,  cos  8 ^  -  cos 2  J.  =  -  2  sin  |(8 ^  +  2  A) sin  |(8 ^  -  2  A) 

=  —  2  sin  5  -4  sin  3  A. 

By  the  converses  of  formulas  (1)  •  •  •  (4),  a  product  involving 
sines  or  cosines  or  both  is  transformed  into  a  sum  or  a  differ- 
ence of  sines  or  cosines. 


IDENTITIES  61 


EXERCISE  XV 

1.  State  in  words  identities  [19]  •  •  •  [22]. 

The  mm  of  the  sines ^       «   .    ^   ,^  .   ,^  ,.«. 

,  ,  ,        !►  =  2  sin  half  sum  •  cos  fialf  difference. 

of  any  two  angles    J 

Prove  each  of  the  following  identities  : 

2.  sin  60°  +  sin  30°  =  2  sin  45°  cos  15°.  by  [19] 

3.  sin  50°  +  sin  10°  =  2  sin  30°  cos  20°. 

4.  cos  75°  +  cos  15°  =  2  cos  45°  cos  30°.  by  [21] 

5.  cos  80°  -  cos  20°  =  -  2  sin  50°  sin  30°. 

6.  sin  7  J.  —  sin  3^  =  2  cos 5^  sin  2^. 

„    sin  7  ^  —  sin  5  ^      2  cos  6  ^  sin  ^ 

7.   =  — =  tanJ.. 

cos  7 .4  4-  cos  o  A      2  cos  6  A  cos  A 

8.  'J^AjLEllA^t^r>2A. 

COS  A  +  cos  3  ^  . 

^     sin  C  +  sin  D       ^      C  +  D       ^  C  -  D       tan  tt  (C  +  D) 

9.   =  tan .  cot = — '-. 

sin  C  —  sin  D  2  2  tan  i  (C  —  D) 

,  -     sin  C  4-  sin  D  C  +  I) 

10.   =  tan 

cos  C  +  cos  D  2 

,,     sinC  +  sinD  C-D  D-C 

11.   =  —  cot =  cot 

cos  C  -  COS  D  2  2 


,„    sin  C  -  sin  D      ^      C-D 

12.   =  tan 

cos  C  +  cos  D  2 

,„     sin  C- sin  D  ^C  +  D 

13.   =  —  cot 

cos  C  —  cos  D  2 

,  ^    cos  (7  +  cos  D  ,  O  +  D         C-D 

14.   =  —  cot cot 

cos  C  -  cos  D  2  2 

=  coti(C  +  D)coti(i)-  C). 

15.  Given  sin  J.  =  1  /2,  sin  B  =  1  /3,  to  find  sin  {A  +  B),  sin  {A  -  B), 
cos  (^4- 5),  cos(^  — J5),  sin  2^,  sin  2  5,  cos  2^,  cos2jB:  (1)  when  A 
and  B  are  both  in  the  first  quadrant ;  (2)  when  A  is  in  the  first  and  B  is 
in  the  second  quadrant. 


62  PLANE  TRIGONOMETRY 

16.  From  the  answers  to  example  15,  find  in  the  simplest  way 
tan  {A  +  B),  tan  {A  -  B),  cot  {A  +  B),  cot  {A  -  B),  sec  {A  +  B), 
CSC  {A  +  B),  tan  2  ^,  cot  2  J.,  sec  2  5,  esc  2  B,  in  cases  (1)  and  (2). 


EXERCISE   XVI 
Examples  for  Review 
Prove  each  of  the  following  identities  : 
1. 


sin  (x  +  y)  _ 

tan  X  +  tan  y 

5. 

„    .      2  -  sec2^ 
cos  2  ^  = 

sin  (x  —  y) 

tan  X  —  tan  y 

sec^^ 

cos  (x  +  y)  _ 

1  -  tan  X  tan  y 

6. 

sec  2  ^  = 

cos  (X  -  ?/) 

1  +  tan  X  tan  y 

csc2 A -2 

cos  {x  +  y)_ 

.    „  ^         2  tan  J. 

i  cot  X  -  tan  y. 

v. 

sin2  J.  =  - -. 

sm  X  cos  y 

1  +  tan2^ 

cos  (x  -  y)  _ 

sin  3  J.  —  sin  A 

■.  tan  X  +  cot  y. 

8. 

=  tan  A 

cos  X  sm  y 

cos  3  J.  +  cos  A 

9.    Express    sin  (3  x/ 4),    cos  (3  x/ 4),    and    tan  (3  x/ 4)    in    terms    of 
cos(3x/2). 

10.  Express  sin  (3  x/ 4),  cos  (3  x/ 4),  and  tan  (3  x/ 4)  in  terms  of  the 
trigonometric  ratios  of  3x/8. 

Prove  each  of  the  following  identities  : 

11.  (sin4  +  cos^)2  =  1 +sin2 J.. 

12.  (sin  A  —  cos  A)'^  =  1  —  sin  2  J.. 

13.  tan  ^  +  cot  ^  =  2  CSC  2  A. 

14.  cot  J.  -  tan  ^  =  2  cot  2  ^. 

tan  ^  +  tan  5 

15.    =  tan  A  tan  B. 

cot  A  +  cot  B 

16.  Given  Sin ^  =2/3,  COS  B=  1/2,  to  find  (1) sin  (^  +  B),  sin(^-B), 
cos{A-\-B),  cos{A-B),  sin2^,  cos2^,  sin2B,  cos2B;  (2)tan(^4--B), 
cot  (A  +  B),  tan  {A  -  B),  cot  {A  -  B),  tan  2  ^,  cot  2  A,  tan  2  J5,  cot  2  B. 

Prove  each  of  the  following  identities : 

,„    cot^  +  cotB  ,„       .,        ,„  ,     .. 

17. =  cos  IB  -  A)  sec  (B  +  A). 

cot  ^  -  cot  5  ^  '       ^  ' 


IDENTITIES  6; 

cos^^  COS  =2  B 

19.  — =  tan  (A  +  B)  tan  (A  —  B). 

l-tan2^tan2^  v  /        v  y 

20.  V2  sin  (A  ±  45°)  =  sin  A  ±  cos  A. 

21.  2  sin  (45°  -  A)  cos  (45°  +  B)  =  cos  (^  -  ^)  -  sin  (^  +  i?). 

22.  2  sin  (45°  +  A)  cos  (45°  +  5)  =  cos  {A  +  B)  +  sin  (^  —  5). 

23.  2  sin  (45°  +  A)  cos  (45°  -  B)  =  cos  {A  -  B) -}-  sin  (^  +  5). 
cot  ^  -  1  11  -  sin  2  ^      1  —  sin  2  ^ 


24.  cot(^+45°)^:::-:_^^-^^ 

p.nt  y4 
25.   cot  (^  -  45°)  = 


cot  J.  +  1       \  1  +  sin  2  ^         cos  2  -4 
cot  ^  +  1      tan  ^  +  1 


1  —  cot  A     tan  A  —  I 

26.  tan  {A  ±  45°)  +  cot  {A  =F  45°)  =0. 

27.  sin9ic  —  sin  7x  =  2cos8xsinx. 

28.  cos  7  X  +  cos  6  X  =  2  cos  6  X  cos  x. 

-  _    sin  3  X  -  sin  X         .  _ 

29.   =  cot  2  X. 

cos  X  —  cos  3  X 

_  -     sin  5  X  —  sin  2  X         ^  7  x 

30.   =  cot 

cos  2  X  —  cos  5  X  2 


31. 


sin  A  -{■  sin  B     cos  ^  +  cos  B 


cos  A  —  cos  B     sin  B  —  sin  A 
32.   tan  (x/ 2  +  45°)  =  tan  x  +  sec  x. 


CHAPTER   IV 
SOLUTION  OF  RIGHT   TRIANGLES   WITH  LOGARITHMS 

41.  In  Chapter  I,  right  triangles  were  solved  without  loga- 
rithms. In  general,  however,  arithmetic  computations  are 
much  abbreviated  by  using  logarithms.  It  is  assumed  that 
the  student  is  already  familiar  with  the  theory  of  logarithms 
from  the  study  of  Algebra  ;  but  to  bring  to  mind  those  proper- 
ties of  logarithms  which  adapt  them  to  shortening  arithmetic 
computations,  a  brief  review  is  given  below. 

42.  Logarithms.  If  a  =  N,  (1) 
then  X,  the  exponent  of  a,  is  called  the  logarithm  of  N  to  the 
base  a,  which  is  written  in  symbols 

x  =  \og,N.  (2) 

Equations  (1)  and  (2)  are  equivalent ;  (2)  is  the  logarithmic 
form  of  writing  the  relation  between  a,  x,  and  N  given  in  (1). 

^.gr.,  since  3^^  =  9, 2  is  the  logarithm  of  9  to  the  base  3  ;  i.e.  +  2  =  logs  9. 
Since  2~^  =  1/8, -3  =  log2(l/8). 

Since  4^^^  =  8, +3/2  =  log4  8. 

-     Ex.  1.     Express  in  the  logarithmic  form  each  of  the  following  relations: 
3*  =  81,  4^  =  64,  6^  =  216,  n<^  =  6,  5"^  =  1/125,  3"^  =  1/243. 
Ex.  2.     Express  in  the  exponential  form  each  of  the  following  relations : 
logs  125  =  3,  log2  32  =  5,  log4  64  =  3, 

\ogcM=h,  log2(l/16)  =  -4. 

Ex.  3.  When  the  base  is  10,  what  is  the  logarithm  of  1  ?  10  ?  100  ? 
1000  ?  10000  ?  100000  ?  0.1  ?  0.01  ?  0.001  ?  0.0001  ?  0.00001  ? 

Ex.  4.  What  is  the  number  when  the  base  is  10  and  the  logarithm 
is0?l?2?3?  -1?   -2?  -3?   -4? 

64 


PROPERTIES  OF  LOGARITHMS  65 

43.  Properties  of  logarithms.  Since  logarithms  are  exponents, 
from  the  general  laws  of  exponents  we  obtain  the  following 
general  properties  of  logarithms  to  any  base. 

(i)  The  logarithm  of  the  product  of  two  or  more  arithmetic 
numbers  is  equal  to  the  sum  of  the  logarithms  of  the  factors. 

Let  JW  =  a^,  N  =  av. 

Then  Mx  N=a^+v. 

Hence  loga(-M'  +  N)  =  x  -{■  y  =  loga-Sf  +  logai^. 

(ii)  The  logarithm  of  the  quotient  of  two  arithmetic  numbers 
is  equal  to  the  logarithm  of  the  dividend  minus  the  logarithm 
of  the  divisor. 

Let  Jf  =  a^,  N  =  aM. 

Then  M  ^  N  =  a^-v. 

Hence  loga  {M  ^  N)  =  x  -  y  =  loga  M  —  loga  JV^. 

(iii)  The  logarithm  of  any  power  of  an  arithmetic  number  is 
equal  to  the  logarithm  of  the  number  multiplied  by  the  'exponent 
of  the  power. 

Let  M  =  a^. 

Then,  for  all  real  values  of  p,  we  have 
M^  =  aP'^. 

Hence  loga  (^^)  =  px  =  p  logaM.  (1) 

If  p  =  1/r,  from  (1)  it  follows  that 

(iv)  The  logarithm  of  any  root  of  an  arithmetic  number  is 
equal  to  the  logarithyn  of  the  number  divided  by  the  index  of  the 
root. 

An  expression  is  said  to  be  adapted  to  logarithmic  computa- 
tion when  it  involves  only  products,  quotients,  powers,  or  roots. 

E.g.,  x'^y^'^ /z^  is  adapted  to  logarithmic  computation;  for  we  have 

loga  (x^y' /'■  /z^)  =  c  \ogaX  +  (1  /r) loga  y  -  8  log„  z.  (1) 

Observe  that  only  the  arithmetic  value  of  a  product,  quotient,  power, 
or  root  is  obtained  by  logarithms ;  the  quality  must  be  determined  by  the 
laws  of  quality. 


66  PLANE  TRIGONOMETRY 

Logarithms  do  not  aid  in  the  operation  of  addition  or  of  subtraction. 
But  when,  as  in  formulas  [19]  •  •  •  [22],  a  sum  or  a  difference  is  identical 
with  a  product,  the  sum  or  difference  can  be  obtained  by  computing  the 
product. 

E.g.,  loga (x2  -  y2)  =  iog„  [(x  +  y)  {x  -  y)']=  loga  (a;  +  2/)  +  loga {x-y). 

44.  Common  logarithms.  The  logarithms  used  for  abridg- 
ing arithmetic  computations  are  those  to  the  base  10 ;  for  this 
reason  logarithms  to  the  base  10  are  called  common  logarithms. 

Thus  the  common  logarithm  of  a  number  answers  the  qyiQS- 
tion,  What  power  of  10  is  the  number? 

IMost  numbers  are  incommensurable  powers  of  10;  hence 
most  common  logarithms  are  incommensurable  numbers,  whose 
approximate  values  we  usually  express  decimally. 

E.g.,  the  common  logarithm 
of  any  number  between  10  and  100  lies  between  +1  and  +2  ; 
of  any  number  between  1  and  10  lies  between  0  and  +1 ; 
of  any  number  between  0. 1  and  1  lies  between  —1  and  0  ; 
of  any  number  between  0.01  and  0.1  lies  between  -2  and  -1 ;  etc. 

Hence  the  common  logarithm 
of  any  number  between  10  and  100  is  +1  +  a  positive  decimal ; 
of  any  number  between  1  and  10  is  0  +  a  positive  decimal ; 
of  any  number  between  0. 1  and  1  is  -1  +  a  positive  decimal ; 
of  any  number  between  0.01  and  0.1  is  -2  +  a  positive  decimal. 

45.  Characteristic  and  mantissa.  A  logarithm  is  said  to  be 
in  the  type  form  when  it  is  expressed  as  the  sum  of  an  integer, 
positive  or  negative,  and  a  positive  decimal  fraction;  in  this 
form  the  integer  is  called  the  characteristic,  and  the  fraction 
the  mantissa. 

In  the  following  pages,  when  no  base  is  written  the  base  10 
is  understood. 

A  negative  characteristic,  as  ~1,  is  usually  written  in  the 
form  1  or  9  —  10;  ~2  in  the  form  2  or  8  —  10;  etc. 


CHARACTERISTICS  67 

The  second  form,  which  is  usually  the  more  convenient  for 
negative  chai-acteristics,  is  sometimes  used  even  when  the 
chai-acteristic  is  positive. 

E.g.,  log 434.1  =  2.63759  ;  +2  is  the  characteristic  and  +.63759  is  the 
mantissa ;  log  0.0769  =  2.88593,  or  8.88593  -  10  ;  -2,  or  8  -  10,  is  the 
characteristic,  and  +.88593  is  the  mantissa. 

In  the  first  form  of  writing  a  negative  characteristic,  the  sign  —  is 
written  above  the  characteristic  to  show  that  this  sign  affects  the  char- 
acteristic only.  One  practical  advantage  of  the  second  form  is  that  we 
can  make  the  positive  part  of  any  logarithm  as  large  as  we  please,  or  the 
negative  part  any  multiple  of  10  we  please. 

E.g.,    log  0.0769  =  2.88593  =    8.88593  -  10  =  18.88593  -  20  =  •  •  -. 
Also,    log 434.1    =  2.63759  =  12.63759  -  10  =  22.63759  -  20  =  •  •  -. 

46.  The  characteristic  of  the  common  logarithm  of  a  number 
is  found  by  the  following  simple  rule : 

Calling  units'  place  the  zeroth  place,  if  the  first  significant 
figure  in  any  number  M  is  in  the  nth  place,  then  the  character- 
istic of  log  M.  is  +  n  or  —  n,  according  as  this  first  figure  is  to 
the  left  or  to  the  right  of  imits'  place. 

E.g.,  when  the  first  significant  figure  of  a  number,  as  5348,  is  in  the 
third  place  to  the  left  of  units'  place,  then  the  number  lies  between  lO^ 
and  10*  ;  hence  its  common  logarithm  is  +3  +  a  mantissa. 

Again,  when  the  first  significant  figure  of  a  number,  as  0.00071,  is  in 
the  fourth  place  to  the  right  of  units'  place,  then  the  number  lies  between 
10—*  and  10— 3 ;  hence  its  common  logarithm  is  — 4  +  a  mantissa. 

Let  the  first  significant  figure  in  the  number  M  be  in  the 
wth  place  to  the  left  of  units'  place ;  then  M  lies  between  lO" 
and  10"  + 1;  that  is, 

IT  __  1  Q«  +  a  positive  decimal. 

.*.  log  3/  =  +71  -f-  a  mantissa. 
Again,  let  the  first  significant  figure  in  the  number  M  be  in 
the  nth  place  to  the  right  of  units'  place ;  then  M  lies  between 
10-«and  lO-e*-^);  that  is, 

T^J  ^  i  Q—  n  +  a  positive  decimal. 

.*.  log  M  =  -71  4-  a  mantissa. 


68 


PLANE   TRIGONOMETRY 


47.  If  the  expressions  of  tivo  numbers  differ  only  in  the  posi- 
tion of  the  decimal  point,  the  two  numbers  have  the  same  mantissa. 

When,  in  the  expression  of  a  number,  a  change  is  made  in 
the  position  of  the  decimal  point,  the  number  is  multiplied  or 
divided  by  some  entire  power  of  10  ;  that  is,  an  integer  is 
added  to  or  subtracted  from  its  logarithm ;  therefore  its  man- 
tissa is  not  changed. 

E.g.,  34.271  x  lO^  =  34271. 

.-.  log  34.271  +  3  =  log  34271.  §  43,  (i) 

Hence  the  mantissa  for  34.271  equals  the  mantissa  for  34271. 


48.    A  convenient  formula  for  computing  the  common  logarithms 

of  whole  numbers  is 

log(.  +  l)=log.  +  2m  (2^  +  3(2/+ 1)'+-)  (=^) 
where  m  =  0.434294,  and  z  is  any  whole  number. 

For  the  proof  of  identity  (a)  see  §  97  in  Taylor's  Calculus 
or  §  322  in  Taylor's  College  Algebra. 

To  compute  log  2,  put  z  =  1  in  (a) ;  to  compute  log  3,  put 
^  =  2 ;  to  compute  log  4,  we  have  log  4  =  2  log  2  ;  to  compute 
log  5,  put  ^  =  4 ;  and  so  on. 

The  series  in  (a)  converges  more  and  more  rapidly  as  z 
increases. 

Note.     Before  proceeding  farther  in  this  chapter,  the  student  should 
familiarize  himself  with  the  use  of  logarithmic  tables,  both  of  natural 
^    numbers  and  of  the  trigonometric  ratios  of 
angles. 

An  explanation  of  the  tables  will  be  found 
in  the  introduction  to  them. 

49.    Right-angled  triangles.     Review 
§13. 

Case  (i).     Ex.  1.      In  the  right  triangle 
ABC,  A  =  48°  17',  and  AB  =  324  ft.;  solve 
Fig.  30  the  triangle. 


RIGHT-ANGLED  TRIANGLES  69 


C  B  =  4\°  43', 
Given  V  ~  o^/'  '    to  find  ^  a  =  241.85, 


i    0  =  324;         t-fi-d|- 


215.6. 

Construct  the  triangle  ABC,  having  the  given  parts. 

r  J5  =  90'  -  ^  =  41°  43', 
I 
Formulas  -i  a  =  c  sin  A, 

I  b  =  c  cos  A. 

.,     .    ,         ,        rioga  =  logc  +  logsin^, 
Logarithmic  formulas    [^^^^^  ^^g ,  +  j^g ^os ^. 

log  c  =  2. 51055  log  c  =  2. 51055 

log  sin  ^  =  9.87300-  10  log  cos  ^  =  9.82311  -10 

.-.  log  a  =  2.38355  .-.  log  b  =  2.33366 

.-.  a  =  241.85.  .-.  b  =  215.6. 

In  Chapter  I  we  checked,  or  verified,  the  calculated  values  by  construc- 
tion and  measurement.  But  these  values  are  more  usually  checked,  or 
tested,  by  using  some  known  relation  between  the  sides  and  angles  which 
has  not  been  employed  in  solving  the  triangle.  Thus,  in  the  example 
above,  we  might  use  either  the  relation  a2  =  c^  -  62  or  tan  ^  =  a/ 6  as  a 
check ;  but  the  former  is  the  better. 

Check.     a2  =  (c  +  b)  (c  -  b).  '      log  (c  +  6)  =  2.73207 

Here  c  +  6  =  539.6,  log  (c  -  6)  =  2.03503 

c- 6  =  108.4.  .-.log  a  =  4.76710/2 

=  2.38355. 
As  this  value  of  log  a  is  the  same  as  that  obtained  in  the  solution  above, 
the  answers  are  probably  correct  to  four  figures. 

Before  using  the  tables  the  student  should  make  a  complete 
outline  of  the  computation  (such  as  he  would  have  by  erasing 
the  second  members  of  the  equations  following  the  logarithmic 
formulas). 

Note  1.  The  direction  above  enables  the  student  to  save  time  by 
writing  at  once  all  the  logarithms  that  are  found  at  one  place  in  the 
table.  Thus  we  find  log  sin  A  and  log  cos  A  at  the  same  time ;  then 
having  both  log  a  and  log  b,  we  next  find  a  and  b. 

Note  2.  When  the  student  has  become  familiar  with  logarithmic 
computations,  he  need  not  write  the  logarithmic  formulas.     By  a  glance 


70 


PLANE  TRIGONOMETRY 


at  the  trigonometric  formulas  he  will  know  how  to  combine  the  logarithms 
in  the  computation  and  can  arrange  his  work  accordingly. 


Note  3.     As  a  check  formula,  we  use  a^  =  (c  -\-  b)  (c 
(c  +  a)  (c  —  a),  according  asc  —  6orc  —  ais  the  greater. 


b)  or  62 


Case  (ii).     Ex.  2.     In  the  right  triangle  ABC,  AB 
=  18.7  ft.  and  CB  =  1G.98  ft.;  solve  the  triangle. 


Given 


rc=  18.7, 
\a  =  16.98; 


to  find 


A  =  65°  14', 
B  =  24°  46', 
b  =  7.8339. 


Fig.  31 
Logarithmic  formulas 


Construct  the  triangle  ABC,  having  the  given  parts. 

sin^  =  a/c, 
Formulas  -I         ^  =  90°-^, 
b  =  a  cot^. 


f  log  sin  J.  =  log  a  —  log  c, 
(^         log  &  =  log  a  +  log  cot  A. 


loga  =  11.22994 
log  c  =    1.27184 


10 


loga  =  1.22994 

log  cot  A  =  9.66404  -  10 

log  sin  A  =    9.95810  -  10  "   .-.  log  b  =  0.89398 

.-.  A  =  65°  14',  .-.  B  =  24°  46'.  .-.  b  =  7.834. 

Check.     a2  =  (c  +  b)  (c  -  6).  log  {c  +  b)  =  1.42380 

Here      c  +  6  =  26.534,  log  (c  -  &)  =  1-03607 

c-6  =  10.866.  .-.loga^  2.45987/2 

=  1.22994. 

As  this  value  of  log  a  is  the  same  as  that  obtained  from  the  table,  the 
answers  are  probably  correct  to  four  places. 


Ex.  3.     Given 


to  find 


{:: 


194.5, 
233.5; 


A  =  39°  47'  36' 
B  =  50°  12'  24' 
c  =  303.9. 


Construct  triangle  ABC,  having  the  given 


parts. 
Formulas 


Fig.  32 


tan^  =  a/6, 
J5  =  90°  -  A, 
c  =  b  sec  A  =  6/cos  A. 


ISOSCELES  TRIANGLES  71 

log  a  =  12.28892  -  10  log  b  =  12.36829  -  10 

log  b  =    2.36829  log  cos  A  =    9.88557  -  10 

.-.  log  tan  A  =    9.92063  -  10  .-.  log  c  =    2.48272 

.-.  A  =  39°  47'  36".  .-.  c  =  303.89. 

.-.  B  =  50°  12'  24". 

Observe  that  the  subtraction  above  is  simplified  by  writing  the  char- 
acteristic 2  of  log  a  and  log  b  in  the  form  12  —  10,  and  the  characteristic 
—  1  of  log  cos  A  in  the  form  9  —  10. 

Check.     b^=  {c  +  a)  {c  -  a).  log  {c  +  a)  =  2.69758 

Here         c  +  a  =  498.4,  log  {c  -  a)  =  2.03902 

c-a  =  109.4.  ••  logft  =  4.73660/2 

=  2.36830. 

As  this  computed  value  of  log  b  differs  by  only  .00001  from  that  found 
in  the  table,  the  computed  parts  are  probably  correct  to  four  places. 


EXERCISE  XVn 

Solve  the  triangle  ABC,  having  given  : 

1.  B  =  67°,  a  =  5.  9.  a  =  3.414,  b  =  2.875. 

2.  A=  38°,  a  =  8.09.  10.  A  =  46°  23',  c  =  5278.6. 

3.  ^  =  15°,  c  =  7.  11.  a  =  529.3,  c  =  902.7. 

4.  5  =  50°,  b  =  20.  12.  B  =  23°  9',  b  =  75.48. 

5.  a  =  0.35,  C  =  0.62.  13.  B  =  18°  38',  c  =  2.5432. 

6.  a  =  273,  b  =  418.  14.  A  =  31°  45',  a  =  48.04. 

7.  b  =  58.6,  c  =  76.3.  15.  b  =  617.57,  c  =  729.59. 

8.  ^  =  9°,  6  =  937.  16.  B  =  82°  6'  18",  a  =  89.32. 

50.  Isosceles  triangles.  In  an  isosceles  triangle  the  perpen- 
dicular from  the  vertex  to  the  base  divides  the  isosceles  triangle 
into  two  equal  right  triangles.  Hence  any  two  parts  which 
determine  one  of  these  right  triangles  determine  also  the 
isosceles  triangle. 


72 


PLANE   TRIGONOMETRY 


In  this  and  the  next  article  we  shall  use  the  following  nota- 


c. 

\ 

tion  in  isosceles  triaDgles  : 

r  =  one  of  the  equal  sides, 
c  =  base, 

V 

h     \ 

h  =  altitude, 

^  =  one  of  the  equal  angles, 
\                        C  =  angle  at  the  vertex. 

/    % 

\                      Q  =  area  of  the  triangle. 

\  7-> 

.  B 

^        Ex.  1.     Given  r  and  c;  to  find  A,  C,  A, 

Fig.  33 

and  Q. 

A=. 

cos-i(c/2r),  0  =  180° -2^, 

h  = 

Vr2  -  (c/2)2  =  V(r  +  c/2)  (r  -  c/2). 

Also,                  h  = 

r  sin  ^  or  7i  =  (c/2)  tan  ^. 

Q  = 

ch/2. 

51.  Regular  polygons.  Lines  drawn  from  the  center  of  a 
regular  polygon  of  n  sides  to  the  vertices  divide  the  polygon 
into  n  equal  isosceles  triangles ;  and  the  perpendiculars  from 
the  center  to  the  sides  of  the  polygon  divide  these  n  equal 
isosceles  triangles  into  2n  equal  right  triangles. 

Hence  any  two  parts  which  determine  one  of  these  equal 
right  triangles  determine  also  the  regular  polygon. 

Using  the  notation  given  in  fig. 
34,  we  have 

C/2  =  3607(2  n)  =  ISOV^i. 

li  p  =  the  perimeter  of  the  poly- 
gon and  F  =  the  area,  we  have 

p  =  nc,  F  =  ph/2. 

CA,  or  r,  is  the  radius  of  the  cir- 
cumscribed circle  and  CD,  or  h,  is 
the  radius  of  the  inscribed  circle. 


REGULAR  POLYGONS  73 


EXERCISE  XVm 

In  an  isosceles  triangle,  having  given : 

1.  c  and  A  ;  find  C,  r,  h. 

2.  h  and  C ;  find  A,  r,  c. 

3.  c  and  ^;  find  A,  C,  r. 

4.  c  =  2.352,  C  =  09°  49' ;  find  r,  A,  ^,  Q. 

5.  h  =  7.4847,  ^  =  76°  14';  find  r,  c,  C,  Q. 

6.  A  barn  is  40  x  80  ft. ,  the  pitch  of  the  roof  is  45° ;  find  the  length 
of  the  rafters  and  the  area  of  both  sides  of  the  roof,  the  horizontal 
projection  of  the  cornice  being  1  ft. 

7.  One  side  of  a  regular  decagon  is  1 ;  find  r,  h^  F. 

8.  The  perimeter  of  a  regular  dodecagon  is  70 ;  find  r,  h,  F. 

9.  In  a  regular  octagon  h=l;  find  r,  c,  F. 

10.  The  area  of  a  regular  heptagon  is  7  ;  find  r,  /i,  p. 

11.  The  side  of  a  regular  octagon  is  24  ft. ;  find  h  and  r ;  also  find  the 
difference  between  the  areas  of  the  octagon  and  the  inscribed  circle,  and 
the  difference  between  the  areas  of  the  octagon  and  the  circumscribed 
circle. 

12.  The  side  of  a  regular  heptagon  is  14  ft. ;  find  the  magnitudes  as 
in  example  11. 

13.  Each  side  of  a  regular  polygon  of  n  sides  is  c ;  show  that  the  radius 
of  the  circumscribed  circle  is  equal  to  (c/2)  esc  (180°/n,),  and  the  radius 
of  the  inscribed  circle  is  equal  to  (c/2)  cot  (180°/n). 

14.  The  radius  of  a  circle  is  k;  show  that  each  side  of  a  regular 
inscribed  polygon  of  n  sides  is  2A:sin  (180°/n),  and  that  each  sido  of  a 
regular  circumscribed  polygon  is  2  fc  tan  (180°/ n). 

15.  The  area  of  a  regular  polygon  of  sixteen  sides  inscribed  in  a  circle 
is  100  sq.  in. ;  find  the  area  of  a  regular  polygon  of  fifteen  sides  inscribed 
in  the  same  circle. 

16.  The  radius  of  a  circle  is  10 ;  find  the  area  between  the  perimeters 
of  two  regular  polygons  of  thirty-six  sides  each,  one  circumscribing  the 
circle  and  the  other  inscribed  in  it. 


CHAPTER   V 


SOLUTION  OF  TRIANGLES  IN  GENERAL 

52.  The  two  following  relations  which  the  sides  of  any 
triangle  bear  to  the  sines  and  cosines  of  its  angles  are  funda- 
mental in  the  study  and  solution  of  triangles.  They  are 
called  the  law  of  sines  and  the  law  of  cosines  respectively. 

Law  of  sines.  The  sides  of  a  triangle  are  proportional  to  the 
sines  of  their  opposite  angles. 


Let  A^  B,  C  denote  the  numerical  measures  of  the  angles  of 
the  triangle  ABC,  and  a,  b,  c  the  numerical  measures  of  its 
sides. 

From  C  draw  CD  _L  BA,  or  BA  produced. 

In  fig.  X,  A  is  acute;  in  fig.  y,  A  is  obtuse. 

In  each  figure  we  have 

by  §7,  ^mB=p/a,  (1) 

by  §  21,  sin  A  =  DC /AC  =p/b.  (2) 

Dividing  the  members  of  (2)  by  those  of  (1),  we  obtain 

sin  A  /sinB  =  a/b,  or  <x/sin  A  =  b  /sin  B.  (3) 

Similarly,  or  by  symmetry  from  (3),  we  obtain 

a/sin  A  =  c/sin  C,    or  b/sin  B  =  c/sin  C. 

74 


LAW  OF  COSINES  76 

Hence  ^  =  ^  =  ^.  [23] 

sin  A       sinB       smC  "■     -^ 

Observe  that  if  C  =  90°,  sin  C  =  1  and  [23]  gives 
a  I  c  =  sin  A  and  h  j  c  =  sin  5, 
which  are  the  known  relations  in  the  right-angled  triangle. 

In  §  62  each  ratio  in  [23]  will  be  shown  to  he  equal  to  the 
diameter  of  the  circle  circumscribed  about  the  triangle  ABC. 

Law  of  cosines.  In  any  triangle  the  square  of  any  side  is 
equal  to  the  sum  of  the  squares  of  the  other  two  sides  minus 
twice  the  ^product  of  these  two  sides  into  the  cosine  of  their 
included  angle. 

In  figures  35  regard  ADj  DB,  and  AB  as  directed  lines. 
Then  in  each  figure  we  have 

AD-\-DB  =  AB;     .',DB  =  c-AD.  (1) 

Squaring  both  members  of  (1)  and  adding  p^j  we  obtain 

(DB'-\-p^)  =  c^  +  (AD^+j)'')-2c-AD.       (2) 

In  each  figure  DB^  -{■p^  =  a^  AD^  -{- p^  =  b% 
and  AD/ AC  =  cos  A. 

Whence  AD  =  b  cos  A. 

Substituting  these  values  in  (2),  we  obtain 

a«  =  b*  +  c*  —  2  be  cos  A.  [24] 

Similarly,  or  by  symmetry  from  [24],  we  obtain 

b^=a-{-c^-2aG  cos  B,  (1) 

c^  =  a-\-b^-2ab  cos  C.  (2) 

Observe  that  if  ^  =  90°,  [24]  becomes  a^  =  b^ -\-  c%  which 
is  the  known  relation  between  the  sides  when  A  =  90°. 
Solving  [24]  for  cos  A,  (1)  for  cos  B,  etc.,  we  obtain 

b^  +  c^  -  a^  „       a^  +  c'*  -  b'*     ^        ^_^^ 

cos  A  = ! — ,   cos  B  = ,  etc.     [25] 

2bc         '  2ac         *  •-     -^ 


^76  PLANE  TRIGONOMETRY 

The  form  of  the  law  of  cosines  in  [25]  is  useful  in  finding 
any  angle  of  a  triangle  from  its  sides. 

53.  If  of  the  six  parts  of  any  triangle  we  have  given  any 
three  (one,  at  least,  being  a  side),  the  triangle,  as  we  know  by 
Geometry,  is  determined  2iTidi  can  be  constructed.  Hence,  in 
the  numerical  solution  of  triangles  by  Trigonometry,  we  must 
consider  the  following /oi^r  cases,  the  given  parts  being: 

(i)    One  side  and  two  angles. 
(ii)   Two  sides  and  the  angle  opposite  one  of  them. 
(iii)   Two  sides  and  their  included  angle. 
(iv)   Three  sides.  ,    ■ 

In  solving  triangles  we  frequently  use  the  two  following 
geometric  properties  of  triangles  : 

I.    The  sum  of  the  three  angles  is  equal  to  180°. 
IT.    The  greater  angle  is  opposite  the  greater  side,  and  vice 
versa. 

54.  Cases  (i)  and  (ii).  If  two  of  the  three  known  parts  of 
a  triangle  are  a  side  and  its  opposite  angle,  a  fourth  part  can 
evidently  be  found  by  the  law  of  sines. 

Hence  cases  (i)  and  (ii)  can  be  solved  by  the  properties  I 
and  II  in  §  53  and  the  law  of  sines. 

(A  =65°, 
Ex.  1.     Given      \^  =  ^0°,     to  find 
[  a  =  50  ; 

Construct  the  triangle  BAC  having  the 
given  parts. 

r  C  =  180°  -{A  +  B)  =  75°, 
Formulas  <    6  =  asin J5/sin^,  (1) 

[  c  =  a  sin  (7/sin  A.  (2) 

''B        Using  the  table  in  §  5,  we  obtain  from  (1) 
Fig.  36  and  (2), 


OBLIQUE-ANGLED  TRIANGLES 


77 


6  =  50  sin  40° /sin  65° 
=  50  X  0.6428/0.9063  =  35.46, 
and  c  =  60  X  0.9659/0.9063  =  53.29. 

Check.     By  construction  and  measurement. 
Ex.  2.     Given         A  =  50°,  C  =  65°,  c  =  30 ;  find  B,  a,  b. 


Ex.  3.     Given 


A  =  60°, 

a  =  3  y/2,   to  find 
b  =  2y/S 


Construct  the  triangle  ABC^  having  the 
given  parts. 

Observe  that  only  one  such  triangle  can 
be  constructed. 

fsin  B  =  6sin^/a,  (1) 

Formulas  i        C  =  IS0°  -  {A  -^  B), 

[        c  =  a  sin  C/sin  A.     (2) 

Since  b<a,  B<A,  i.e.  B< 60°. 


Fig.  37 


From  (1), 
Hence,  by  §  10, 


sin  B 


3  V2  2 


B  =  45°,  since  B  <  60°. 

C  =  180°  -{A-hB)=  180°  -  105°  =  75°. 
From  (2),  c  =  3  V2  x  0.9659  -  ( V3/2)  =  4.73. 

Check.     By  construction  and  measurement. 
Ex.  4.     II  C  =  60°,  a  =  2,  c  =  V6,  find  6,  ^,  B. 
Ex.  5.     If  ^  =  30°,  a  =  9,  6  =  6,  find  B,  C,  c,  having  given 
sin  19°  28'  =  1/3  and  sin  130°  32'  =  0.76. 

55.  Cases  (iii)  and  (iv).  If  two  sides  of  a  triangle  and  their 
included  angle  are  known,  the  third  side  can  be  found  by  [24]  ; 
if  the  three  sides  are  known,  each  angle  can  be  found  by  [25]. 
Hence  cases  (iii)  and  (iv)  can  be  solved  by  the  law  of  cosines. 

f^  =  60°,  ra  =  7, 

6  =  8,      to  find  J  jB  =  cos-i (1  /7), 
c  =  5;  I  C  =  cos- 1(11/ 14). 


Ex.  1.     Given 


78  PLANE  TRIGONOMETRY 

r       a2  =  62  _i_  c2  -  2  6c  cos^,  (1) 

Formulas  ^  cos  ^  =  {a^  +  c^  -  6^)  /  (2  ac),  (2) 

t  cos  C  =  (a2  +  62  -  c2)  /  (2  ah).  (3) 

From  (1),  a2  =  82  +  52  -  2  •  8  •  5  •  (1/2)  =  49. 

From  (2),  cos  J5  =  (72  +  52  -  82) /(2  •  7  •  5)  =  1/7. 

From  (3),  cos  C  ==  (72  +  82  -  52)/(2  •  7  •  8)  =  11/14. 

.-.a  =  7,     5  =  cos-i(l/7),     0  =  cos- 1(1 1/14). 

Check.     By  construction  and  measurement. 

Ex.  2.     If  a  =  7,  6  =  3,  c  =  5,  find  A.,  B,  O,  having  given 
11  / 14  =  cos  38°  13',  13/ 14  =  cos  2P  47'. 

62  +  c2  -  a2      32  +  52-72  1 

cos  A  = = = . 

2  6c  2.3.5  2 

„      a2  +  c2  -  62      72  +  52  -  32      13 

cos  B  = = =  — 

2  ac  2.7.5  14 

a2  +  62  -  c2      72  +  32  -  52      11 

cos  C  = = =  —  • 

2  a6  2.7.3  14 

.-.  A  =  120°,     B  =  21°  47',     C  =  38°  13'. 
Check.     A  +  B+C  =  120°  +  21°  47'  +  38°  13'  =  180°. 

Ex.  3.     If  a  =  2,  6  =  3,  c  =  4,  find  A,  B,  C,  having  given 

7  /8  =  cos  28°  57',  11  / 16  =  cos  46°  34',  1/4  =  cos  75°  31'. 

56.  Since  the  law  of  cosines  involves  sums,  it  is  not  adapted 
to  computation  by  logarithms.  Hence,  to  solve  cases  (iii)  and 
(iv)  by  logarithms,  we  must  deduce  other  formulas,  one  of 
which  is  the  law  of  tangents  below. 

Law  of  tangents.  I'he  sum  of  any  two  sides  of  a  triangle  is 
to  their  difference  as  the  tangent  of  half  the  sum  of  their  oppo- 
site angles  is  to  the  tangent  of  half  their  difference. 

From  the  law  of  sines,  we  have 

a/h  —  sin  A  /sin  B.  (1) 


LAW  OF  TANGENTS  79 

By  principles  of  proportion,  from  (1),  we  obtain 

a  -h  b  _  sin  A  -\-  sin  B 
a  —  b      sin  A  —  sin  B 

-  2  cos  H^  +  B)  sin  H^  -  B)       ^^  ^1^].  L^^] 
=  tan  i  (A  +  B)/tan  i  (A  -  B).  [26] 

Since  tan  ^  (A  +  B)  =  tan  i  (180°  -  C) 

=  tan  (90°  -  C/2)  =  cot  (C/2), 
from  [26^  we  obtain 

A  —  B       a  —  b       C  roT-i 

tan = cot  - .  [27] 

2  a  +  b       2  •-     -■ 

As  a  check,   [26]  is  the  more  convenient  form,  while  for  solving 
triangles,  [27]  is  the  preferable  form  of  this  law. 

Example.     If  a  =  V3,  &  =  1,  C  =  30°,  find  A,  B,  c,  having  given 
cot  15°=  (V3+  1)/(V3-1). 

^    r^^,.  A-  B      a-b        C 

By  [27],  tan = cot  - 

•^  '•    ■"  2  a  +  6        2 

=  ^^~^cotl5°=l. 
V3  +  1 

Hence  i(^-^)=45°.  (1) 

Also,  1{A  +  B)  =  l  (180°  -  C7)  =  75°  (2) 

Adding  (1)  and  (2),  A  =  120°. 

Subtracting  (1)  from  (2),       B  =  30°. 

Since  C  =  B,  we  have  c  =  b  =  1. 

Check.     By  construction  and  measurement. 


SOLUTION  OF  OBLIQUE  TKIANGLES  WITH  LOGARITHMS 

57.    Case  (i).     Given  one  side  and  two  angles. 

r  a  =  180,  r  C  =  66°  17', 

Ex.  1.     Given     \a  =  38°,         to  find  ^    b  =  283.33, 

5  =  75°  43' ;  [  c  =  267.68. 


80 


PLANE  TRIGONOMETRY 


Construct  triangle  ABC,  having  the  given  parts. 


Formulas 


Logarithmic  formulas 


r  C  =  180°  -  (J.  +  J5)  =  66°  17', 
\    i>  =  a  sin  B / sin  A, 
I  c  =  asin  C/sinA. 

flogb  =  log  a  +  log  sin  B  —  log  sin  ^, 
\  log  c  =  log  a  +  log  sin  G  —  log  sin  A. 


Fig.  38 


loga=    2.25527 
log  sin  B  =    9.98636  -  10 
12.24163  -  10 
log  sin  A  =    9.78934  -  10 
.-.  log  b  =    2.45229 

.-.  b  =  283.33 


Check. 


b  +  a  _t3in^{B  +  A) 
b  —  a     tan  1{B  —  A) 

b  +  a  =  463.33. 
b-a  =  103.33. 


loga  = 

log  sin  C  = 

12.21695 

log  sin  A  =    9.78934 

.-.  logc=    2.42761 

.-.  c  =  267.68 


2.25527 

9.96168  -  10 
10 
10 


log(6  + a)  =  2.66589 

log(6-a)  =  2.01423 

log  quotient  =    .65166 


(5  +  ^)/2  =  56°51'30'^ 
(5-^)/2  =  18°5r30^ 

log  tan  1{B  +  A)  =  10. 18514  -  10 

logtan  ^{B-A)=    9.53347  -  10 

log  quotient  =      .65167 


(1) 


As  the  logarithms  of  the  two  members  of  (1)  differ  by  only  1  in  the 
fifth  place,  the  value  of  b  is  correct  to  four  places. 
Similarly  we  can  check  the  value  of  c. 


OBLIQUE  TRIANGLES  81 


EXERCISE  XTX 

Solve  each  of  the  following  triangles : 

1.    Given  B  =  60°  15', 

C  =  54°  30', 

a  =  100. 

2.   Given  ^  =  45°  4r, 

C  =  62°  5', 

6  =  100. 

3.    Given  B  =  70°  30', 

C  =  78°  10', 

a  =  102. 

4.    Given  A  =  65°, 

B  =  65°, 

c  =  270. 

5.    Given  a  =  123, 

B  =  29°  17', 

C  =  135° 

6.   Given   6=1006.62, 

A  =  44°, 

C  =  70°. 

7.  A  ship  S  can  be  seen  from  each  of  two  points  A  and  B  on  the 
shore.  By  measurement,  AB  =  800  ft.,  Z  SAB  =  67°  43',  and  Z  SBA 
=  74°  21'  16".     Find  the  distance  of  the  ship  from  A. 

8.  A  flag  pole  A  is  observed  from  two  points  B  and  C,  1863  ft. 
apart.  Given  Z  BCA  =  36°  43'  and  Z  CBA  =  57°  21',  find  the  distance 
of  the  flag  pole  from  the  nearer  point. 

9.  To  determine  the  distance  of  a  hostile  fort  A  from  a  place  B,  a 
line  BC  and  the  angles  ABC  and  BCA  were  measured  and  found  to  be 
322.55  yd.,  60°  34',  and  56°  10'  respectively.     Find  the  distance  AB. 

10.  A  balloon  is  directly  over  a  straight  level  road,  and  between  two 
points  on  the  road  from  which  it  is  observed.  The  points  are  15847  ft. 
apart,  and  the  angles  of  elevation  are  found  to  be  49°  12'  and  53°  29' 
respectively.  Find  the  distance  of  the  balloon  from  each  point  of 
observation. 

11.  To  find  the  distance  from  a  point  ^  to  a  point  B  on  the  opposite 
side  of  a  river,  a  line  AC  and  the  angles  CAB  and  ACB  were  measured 
and  found  to  be  316.32  ft.,  68°  43',  and  57°  13'  respectively.  Find  the 
distance  AB. 

12.  From  points  A  and  B,  at  the  bow  and  stern  of  a  ship  respectively, 
the  foremast,  C,  of  another  ship  is  observed.  The  points  A  and  B  are 
300  ft.  apart;  the  angles  ABC  and  BAC  are  found  to  be  65°  31'  and 
110°  46'  respectively.  What  is  the  distance  between  the  points  A  and  C 
of  the  two  ships  ? 


82 


PLANE  TRIGONOMETRY 


58.    Case  (ii).     Given  two  sides  and  an  angle  opposite  one  of 

them.     Let  a,  b,  A  be  the  given  parts.     Then,  to  find  B,  C,  c, 

we  have  -     ^      7    ■     .  , 

sinB  =  osinA/a,  (1) 

C  =  180°  -(A  -\-  B), 

c  =  a  sinC/sin  .4. 

Since  two  supplementary  angles  have  the  same  sine  (§  30), 
the  relation  in  (1)  gives  in  general  two  values  for  B,  both  of 
which  are  to  be  taken  unless  one  is  excluded  by  the  conditions 
of  the  problem. 

We  have  to  consider  the  three  following  cases : 

I,  when  a>  h\  II,  when  a  =  h\  III,  when  a  <  b. 

I.  When  a  >  b  and  A  is  acute  or  obtuse,  then  A  >  B ;  hence 
B  is  acute,  and  there  is  but  07ie  triangle  having  the  given  parts. 

II.  When  a  =  b  and  A  is  acute,  then  A  =  B  ;  hence  B  is  acute 
and  the  triangle  is  isosceles. 

In  this  case  the  triangle  can  be  solved  by  the  method  in  §  50  or  by  the 
law  of  sines  and  the  relation  A  +  B  -\-  C  =  180°. 

If  a  =  6  and  A  =  or  >  90°,  the  triangle  is  impossible.     Why  ? 

Note.  Example  1  below  and  the  first  three  examples  in  Exercise  XX 
may  be  solved  before  Case  III  is  considered. 


Bi  J)         By 


^  A 


y 

Q 

V 

a 

A     rn 

B  ., 

Fig.  39 


III.  When  a  <  b  and  A  is  acute,  there  are  two  triangles,  one, 
or  no  triangle,  having  the  given  parts,  according  as  &">,==,  or 
<  b  sin  A. 

Geometric  proof.    In  each  figure,  let  Z  XA  C  =  A  and  AC  =  b. 


OBLIQUE  TRIANGLES  83 

Draw  CD  A.  AX]  then  in  each  figure  CD  =  b  sin  A. 

With  C  as  a  center  and  a  as  a  radius,  describe  the  arc  mn. 

li  a  >  b  sin  A  (i.e.  if  a  >  CD),  the  arc  mn  will  cut  AX 
(fig.  x)  in  two  points,  Bi  and  B2,  on  the  side  of  A  toward  D, 
and  there  will  be  two  unequal  triangles  having  the  parts  a,  b, 
A,  viz.,  the  triangles  ACB^  and  ACB^.  Hence  B  has  the  two 
values  Z.  AB]^C  and  Z  AB^C,  which  are  supplementary. 

If  a,  =  ^  sin .4,  the  arc  mn  will  touch  AX  at  D  (fig.  y); 
hence  B  =  90°  and  only  the  right-angled  triangle  A  CD  has 
the  given  parts. 

If  a  <  b  sin  ^,  the  arc  mn  will  not  meet  AX  (fig.  z) ;  hence 
no  triangle  can  be  constructed  with  the  given  parts. 

E.g.,  if  a  =  5,  6  =  7,  and  A  =  30°,  then  a<b  and  a  >  6  sin  J. ;  hence 
there  will  be  two  triangles  having  these  parts. 

If  a  <  6  and  ^  =  or  >  90°,  the  triangle  is  impossible.     Why  ? 

Trigonometric  proof.     From  the  law  of  sines, 

sin  B  =  b  sin  A /a,  or  (b  /a)  sin  A.  (1) 

Also  ^  <  5  and  ^  -f  ^  <  180°.  (2) 

If  a  >  ^  sin  A,  b  sin  A  /a  <  1,  whence  from  (1),  sin  5  <  1 ; 
hence  B  has  two  unequal  values,  which  are  supplementary. 

Since  a<b,  b/a>l,  whence  from  (1),  sin  B  >  sin  A  ;  hence 
each  of  the  two  supplementary  values  of  B  will  satisfy  both 
the  conditions  in  (2). 

Therefore  B  has  two  values  and  there  are  two  different 
triangles  having  the  given  parts  (fig.  x). 

If  a  =  b  sin  A,  sin  J5  =  1  in  (1),  whence  B  =  90° ;  hence  the 
required  triangle  is  right  angled  at  B  (fig.  y). 

If  a<bsinA,  sin  B  >  1,  which  is  impossible;  hence  the 
triangle  is  impossible  (fig.  z). 

From  the  trigonometric  proof,  it  follows  that  if  a<b  and  A 
is  acute,  there  are  two  triangles,  one,  or  no  triangle,  according 
as  log  sin  B  is  negative,  zero,  or  positive.     Why  ? 


84 


PLANE  TRIGONOMETRY 


Ex.  1.     Given 


to  find 


a  =  250, 
b  =  240, 
A  =  72°  4' ; 

B  =  65°  58'  24'', 
C  =  41°  57'  36", 
c  =  175.69. 


Here  a>b  and  ^  < 90° ;  hence  B < 90°  and  there 
£  is  only  one  triangle  having  the  given  parts. 
Fig.  «)  Construct  the  triangle  ABC,  having  the  given 

parts. 

sin  B  =  bsmA/a, 
Formulas     •{        C  =  180°  -  {A -\-  B), 
c  =  asm  C /sin  A. 


log  6=    2.38021 
log  sin  J.  =    9.97837  - 
12.35858  - 

loga=    2.39794 
log  sin  B=    9.96064- 

10 
10 

■10 

loga=    2.39794 
log  sin  C=    9.82517-10 
12.22311  -  10 
log  sin  ^=    9.97837  -10 

log  c  =    2.24474 

.-.  B  =  65°  58'  24" 
C  =  180°  -  (72^ 

.-.  c  =  175.69. 
>  4'  +  65°  58'  24")  =  41°  57'  36". 

Check. 

6  +  c      tan  i  (5  +  0) 
b-c     tan  1(5-0) 

( 

Here 

b  +  c  =  415.69, 
b-c  =  64.31, 

i(B+  C)  =  53°58'; 
i(B-C)  =  12°0'24". 

log  (6 +  c)  =  2.61877 
log  {b-c)  =  1.80828 

log  tan  i  (B  +  C)  =  10. 13821  -  10 
log  tan  1(5  -  0)  =    9.32772  -  10 

(1) 


log  quotients    .81049 


log  quotients      .81049 


As  the  logarithms  of  the  two  members  of  (1)  are  equal,  the  values 
obtained  above  are  correct. 

Ex.  2.     How  many  triangles  are  there  which  have  the  following  parts  ? 
(i)  a  -I  70,  6  =  90,  A  =  30°. 


(ii)  a  =  40, 

6  =  80, 

A  =  30°. 

(iii)  a  =  20, 

6  =  50, 

A  =  30° 

(iv)  a  =  70, 

6  =  75, 

A  =  60°. 

OBLIQUE  TRUNGLES 


85 


Ex.  3.     Given 


a  =  732, 

b  =  1015,  to  find 
A  =  40°: 


B  =  63°  2'  20"  or  116°  57'  40" 
C  =  16°  57'  40"  or  23°  2'  20" 
c  =  1109.4  or  445.66. 


Here  a<b  and  a>b  sin  A;  hence  by  III  there  are  two  solutions. 
Construct  the  two  triangles  A  CBi  and  A  CB2,  having  the  given  parts. 


Formula  for  B. 


sin  B  =  bsinA/a. 
log  6=    3.00647 
log  sin  A  =    9.80807  - 


10 


log  a 


12.81454 
2.86451 


10 


.-.log  sin  ^=    9.95003-10 
.CBi  =  63°  2'  20",     B2  =  116°  57'  40".-^ 
e-- jL 


To  find  the  unknown  parts  of  At^i,  we  have 

ZACBi  =  180°  -{A  +  Bi)  =  76°  67'  40" 
ABi  =  a  sin  ^  CBi/sin  A. 

.0 


Fig.  41 

To  find  the  unknown  parts  of  A  CB2,  we  have 

ZACB2  =  180°  -{A  +  B2)  =  23°  2'  20". 
AB2  =  a  sin  ^  C-S2/sin  A. 


\oga=    2.86451 
log  sin  A  CBi  =    9.98866  -  10 


Check. 


12.85317  - 

10 

log  sin  A 

=    9.80807  - 

10 

.'AogABi 

=    3.04510 

.-.  ABi 

=  1109.4. 

c  +  b 

tani(C  +  J5) 

c-b 

tan  1(0  -B) 

loga=    2.86451 
log  sin  A  CB2  =    9.59257  -  10 
12.45708  -  10 
log  sin  A  =    9.80807  -  10 
.-.log  ^^2=    2.64901 
.-.  ABi  =  445.66. 


(1) 


86 


PLANE  TRIGONOMETRY 


Here  c  +  6  =  2124.4, 

c-b=     94.4, 

log  (c  + 6)  =  3.32723 

log  {c-b)  =  1.97497 

log  quotient  =  1.35226 


1 

log  tan  1  (C  +  2^)  =  10.43893  -  10 

log  tan  U<^-  ^)  =    9.08670  -  10 

log  quotients    1.35223 


The  equality  of  these  logarithms  to  five  figures  verifies  the  answers  to 
four  figures. 

EXERCISE  XX 

Solve  the  following  triangles : 

1.  Given  a  =  145,  6  =  178, 

2.  Given  h  =  573,  c  =  394, 

3.  Given  a  =  5.98,  6  =  3.59, 

4.  Given  a  =  140.5,  6  =  170.6, 
0  5.    Given  6  =  74.1,  c  =  64.2, 
r    6.   Given  a  =  27.89,        ^  6  =  22.71, 

^.—  7.    Given  6  =  45.21,  c  =  50.3, 

8.  Given  a  =  34,  6  =  22, 

9.  Given  a  =  55.55,  6  =  66.66, 
YUO.    Given  a  =  309,               6  =  360, 


B=    41°  10'. 
B  =  112°  4'. 
A  =    63°  50'. 
A=    40°. 


C=    27°  18'.' 4?  :   ^.39-^6 


11.   Given  6  =  19,  c  =  18, 


5=  65°  38'. 

B=  40°  32' 7". 

B=  30°  20'. 

5=  77°  44' 40". 

A=  21°  14' 25". 

C=  15°  49'. 


59.    Case  (iii).      Given  two   sides  and  their  included   angle. 

Let  a,  b,  C  be  the  given  parts.     Then,  to  find  A,  B,  c,  we  have 


tan 


7  cot  —y 

a  +  b        2 


(1) 


(^+jB)/2  =  90°-  C/2,  (2) 

c  =  a  sin  C/sin  ^.  (3) 

From  (1)  we  obtain  {A  —  B)/%     Having  given  {A—B)/2 
and  {A  +  5)/2,  we  readily  obtain  A  and  B. 
Then  c  can  be  found  by  (3). 


(a 


Example.   Given 


=  540, 
6  =  420, 
C=52°6'; 


to  find 


r^  =  78°17'40", 
49°  36'  20", 
435.15. 


1" 


OBLIQUE  TRIANGLES 


87 


Formulas 


tan 


B     a 


Here 


6        C 

cot  — , 

a  +  b       2 

(A H- 5)/2  =  90°  -  C/2  =  63°  57', 

c  =  asin  C/sinA. 

a  +  6  =  960,  a-b=  120,   C/2  =  26°  3'. 


log  (a  -  b) 
log  cot  (C/2) 


2.07918 
0.31086 


12.39004 
2.98227 


10 


log  (a  +  6) 
,-.  log  tan  ^  (^  -  ^)  =    9.40777  -  10 

.-.  {A  -5)/2  =  14°20'40". 
Also      (^  +  J5)/2  =  63°67'. 


loga=    2.73239 

log  sin  C  =    9.89712  -  10 

12.62961  -  10 

log  sin  A  =    9.99087  -  10 

.-.  log  c  =    2.63864 

.•.c  =  436.15. 


.-.  A  =  78°  17'  40", 
and  B  =  49°  36'  20". 

Check.     ^  +  B  +  C  =  52°  6'  +  78°  17'  40"  +  49° 
To  check  c  we  could  use  c  =  6  sin  C/sin  B. 


20"  =  180°. 


EXERCISE  XXI 

Solve  each  of  the  following  triangles  : 


1. 


Given  a  =  266, 
Given  b  =  91.7, 
Given  a  =  960, 
Given  a  =  886, 
Given  b  =  41.02, 


b  =  362, 
c  =  31.2, 
b  =  720, 
b  =  747, 
c  =  45.49, 


C  =  73°.         d  t  1(^-3'*'^^ 
^  =  33°7'9".i^-''^2'-'^''*'^ 
C  =  25°  40'. 
C  =  71°  54'.  A^  ^  *^  i*^  ? 


62<'9'38".^-/^-;;;i,^ 
The  dis- 


3. 

4. 

6.  Two  trees  A  and  B  are  on  opposite  sides  of  a  pond 
tance  of  A  from  a  point  C  is  297.6  ft.,  the  distance  of  B  from  C  is 
864.4  ft.,  the  angle  AC  Bis  87°  43'  12".     Find  the  distance  AB. 

7.  Two  mountains  A  and  B  are  respectively  9  and  13  miles  from  a 
town  C,  and  the  angle  ACB  is  71°  36'  37".  Find  the  distance  between 
the  mountains. 

8.  Two  points  A  and  B  are  visible  from  a  third  point  C,  but  not 
from  each  other.  The  distances  AC,  BC,  and  the  angle  ACB  were 
measured  and  found  to  be  1321  ft.,  1287  ft.,  and  61°  22' -respectively. 
Find  the  distance  AB. 

9.  From  a  point  3  mi.  from  one  end  of  an  island  and  7  mi.  from  the 
other  end,  the  island  subtends  an  angle  of  33°  55'  15".  Find  the  length 
of  the  island. 


88  PLANE  TRIGONOMETRY 

10.  Two  stations  A  and  B  on  opposite  sides  of  a  mountain  are  both 
visible  from  a  third  station  C.  The  distance  AC,  BC,  and  the  angle 
ACB  were  measured  and  found  to  be  11.5  mi.,  9.4  mi.,  and  59°  31' 
respectively.     Find  the  distance  from  A  to  B. 

11.  Two  trains  leave  the  same  station  at  the  same  time  on  straight 
tracks  that  form  an  angle  of  21°  12'.  Their  average  speeds  are  40  mi. 
and  50  mi.  an  hour  respectively.  How  far  apart  will  they  be  at  the  end 
of  the  first  thirty  minutes  ? 

60.  Case  (iv).  Given  the  three  sides.  To  solve  this  case  we 
first  obtain  formulas  for  the  sine,  cosine,  and  tangent  of  A  /2, 
B /2,  and  C/2  in  terms  of  the  three  sides  a,  h,  c. 

Let  2s  =  a-{-b  +  c. 

Then  2{s  —  a)  =  h  +  c  —  a, 

2  (s  —  b)  =  a  -\-  c  —  b, 

2  (s  —  c)  =  a  -\-  b  —  c. 

By  [16]  and  [17],  we  have 

sin2(^/2)  =  (l  -cos^)/2, 

cos^(A/2)  =(1  +  cos  A)/2. 
Substituting  for  cos  A  its  value  given  in  [25],  we  obtain 

^^^2=2^^ 2b^)     ^"^   2==2V^  +  ^2^ 

4:  be  4:  be 

_(a  —  b-\-c)(a-\-b-c)  _(b -hc  +  a)(b-^c-a) 

~  Abe  4  be 


(1) 


4:(s  —  b)  (s  —  c)  _4:S  (s  —  a) 

Abe  Abe 


A  (s-b)(s-c)  A  s(s-a) 

.•.sin-  =  \^ P -.        .•.cos-  =  A-  ^ 

2        \  be  2        \ 

-b)(8- 

s(s  —  a) 


be 

^.   .-,.  A         /(s  _  b)(s  -  c) 

Dmdmg,  ^''°2=V      s(s  -  a)       ' 


by(l) 

K28] 


THREE  SIDES  GIVEN  89 

Observe  that  s  is  the  half -perimeter,  a  the  side  opposite,  and  b  and  c 
the  sides  including  the  angle  A, 

Ex.  1.     State  in  words  the  relations  in  [28]. 

Ex.  2.     Write  the  values  of  the  sine,  cosine,  and  tangent  of  5/2;  C/2. 

Since  the  trigonometric  ratios  of  A/ 2,  B/2,  C/2  must  all 
be  positive  (?),  only  the  positive  roots  in  [28]  are  taken. 

Formulas  [28]  enable  us  to  obtain  any  angle  of  a  triangle 
from  the  sides,  either  through  its  sine,  its  cosine,  or  its  tangent. 

Ex.  3.  If  a  =  13,  6  =  14,  c  =  15,  find  A,  B,  C,  having  given 
1/2  =  tan  26°  34',  4/7  =  tan  29°  45',  2/3  =  tan  33°  41'. 

Here  s  =  (13  +  14  +  15)/2  =  21,      s  -  a  =  21  -  13  =  8, 

s  -  6  =  21  -  14  =  7,  s  -  c  =  6. 

Hence  tan  |  =  yj^^  =  1  =  tan  26°  34'. 

.-.  A/2  =  26°  34',  or  ^  =  53°  8'. 

Again,     ~~~^~^^tan  —  =  -\  / =  -  =  tan  29°  45'. 

^      '  ^^^2       \21-7      7 

.-.  B/2  =  29°  45',  or  B  =  59°  30'. 

Similarly  ^  C/2  =  33°  41',  or  C  =  67°  22'. 

Check.     ^  +  5  +  C  =  53°  8'  +  59°  30'  +  67°  22'  =  180°. 

Ex.  4.  If  a  =  35,  6  =  84,  c  =  91,  find  A,  B,  C,  having  given 
tan  11°  19'  =  1/5,  and  tan  33°  41'  =  2/3. 

Ex.  5.     If  a  =  13,  6  =  14,  c  =  15,  find  the  sines  of  ^/ 2,  5/2,  0/2. 

If  we  multiply  the  value  of  tan(.l  /2)  in  [28]  by  1  in  the 
form  Vs  —  a  /  Vs  —  a,  the  expression  for  tan  (.1/2)  can  be 
put  in  the  more  symmetrical  form 

tan:^  =  -l-     /(s-a)(8-b)(s-c) 
2       s  —  a  \ 8 

Letting  r  =  ^(s  -  a)  (s  -  b)  (s  -  c)/s,         [29] 

we  have  tan(A/2)  =  r/(s  —  a).^ 

Similarly  tan  (B  /  2)  =  r /  (s  -  b),  I  [30] 

tan  (C/2)  =  r/(s-c).J 


90 


PLANE  TRIGONOMETRY 


When  all  the  angles  are  required,  the  tangent  formulas  [30] 
are  the  best,  as  they  involve  the  fewest  logarithms. 

In  §  63  it  will  be  proved  that  r  in  [29]  is  the  radius  of  the 
circle  inscribed  in  the  triangle  ABC. 


Ex.  6.     Given 


Formulas 


a  =  130, 

6  =  123,  to  find 

c  =77; 


A  =  77°  19'  9", 
B  =  67°  22'  49'', 
C  =  36°  18'  2". 


r  =  V(.s  —  a)  (s  —  b)  (s  —  c)/s, 
tan  (A  /2)  z=  r / (s  —  a), 
tan  (5/2)  =r/{s-  6), 
tan(C/2)  =  r/{s-  c). 


Here  s  =  165,  s 

\og{s-a)=  1.54407 

log(s  -b)  =  1.62325 

log  {s-  c)  =  1.94448 

5.11180 

log  s  =  2.21748 

.-.  log  r  =  2.89432/2 


a  =  35,  s  -  6  =  42,  s  -  c  =  88. 

.-.  log  tan  {A/2)  =  9.90309  -  10, 
whence  ^/2  =  38°  39'  34.6" ; 

log  tan  (B/2)  =  9.82.391  -  10, 
whence  B/2  =  33°  41'  24.4"; 

log  tan  (C/2)  =  9.50268  -  10, 
whence  C/2  =  17°  39'  1". 


=  11.44716  -  10. 
Check.     A  +  B+C  =  11°W  9"  - 


67°  22'  49"  +  35°  18'  2"  =  180^ 


EXERCISE   XXII 

Solve  each  of  the  following  triangles : 

1.    Given  a  =  56,                         b  =  43, 

c:^49. 

2.    Given  a  =  8.5,                        6  =  9.2, 

c  =  7.8. 

3.    Given  a  =  61.3,                      b  =  84.7, 

c  =  47.6. 

4.    Given  a  =  705,                       b  =  562, 

c  =  639. 

5.    Given  a  =  .0291,                    b  =  .0184, 

c  =  .0.358. 

6.    Given  a  =  85,                         b  =  127, 

A  =  26°  26'. 

7.    Given  a  =  5.953,                    b  =  9.639, 

C  =  134°. 

8.    Given  a  =  3019,                     b  =  6731, 

c  =  4,228. 

9.    Given  a  =  60.935,                  c  =  76.097, 

A  =  133°  41' 

THREE  SIDES  GIVEN  91 

10.  Given  h  =  74.8067,  c  =  98.738,  C  =  81°  47'. 

11.  Given  b  =  129.21,  c  =  28.63,  A  =  27°  13'. 

12.  Given  a  =  2.51,  6  =  2.79,  c  =  2.33. 

13.  Given  a  =  32.163,  c  =  27.083,  C  =  52°  24'  16". 
•     14.  Given  a  =  74.8,  c  =  124.09,  B  =  83°  26'  52". 

15.  Given  a  =  86.0619,  c  =  63.5761,  A  =  19°  12'  43". 

16.  Given  a  =  93.272,  b  =  81.512,  C  =  58°. 

17.  The  sides  of  a  triangular  field  are  534  ft.,  679.47  ft.,  and  474.5  ft. 
Find  the  angles. 

18.  A  pole  13  ft.  long  is  placed  6  ft.  from  the  base  of  an  embankment, 
and  reaches  8  ft.  up  its  face.     Find  the  slope  of  the  embankment. 

19.  A  point  P  is  13581  in.  from  one  end  of  a  wall  12342  in.  long, 
and  10025  in.  from  the  other  end.  What  angle  does  the  wall  subtend  at 
the  point  P  ? 

20.  The  distances  between  three  cities  A,  B,  and  C  are  as  follows : 
^J5  =  165  mi.,  .4C  =  72  mi.,  and  BC  =  185  mi.  B  is  due  east  from  A. 
In  what  direction  is  C  from  A? 

21.  Under  what  visual  angle  is  an  object  7  ft.  long  seen  when  the  eye 
of  the  observer  is  5  ft.  from  one  end  of  the  object  and  8  ft.  from  the 
other  end  ? 

22.  Prove  sin  ^  =  2  Vs  (s  -  a)  {s  -  b)  (s  -  c)  /  (6c). 

23.  Prove  cos^  =  s (s  -  a) /(6c)  -  (s  -  6)  (s  -  c) /(6c). 

24.  If  a  =  18,  6  =  24,  c  =  30,  show  that  sin ^  =  3/5. 

25.  The  sides  of  a  triangle  can  be  substituted  for  the  sines  of  their 
opposite  angles,  and  vice  versa,  when  they  are  involved  homogeneously  in 
the  numerator  and  denominator  of  a  fraction,  or  in  both  members  of  an 
equation. 

For  this  substitution  is  the  multiplication  of  each  term  by  the  equal 
numbers,  a /sin  A,  6 /sin  jB,  c/sin  C,  or  their  reciprocals. 

E.g.,  sin  A  =  sin  (180°  -  A)  =  s\n{B-\-  C). 

.'.  sin  A  =smB  cos  C  +  sin  O  cos  B.  by  [7] 

Multiplying  the  first  term  hy  a/sinA,  the  second  by  6 /sin  B,  and  the 
third  by  c/sin  C,  we  obtain 


92  PLANE  TRIGONOMETRY 

a  =  b  cos  C  +  c  cos  B.   ^ 
Similarly  b  =  a  cos  C  +  c  cos  A,  ^  (1) 

and  c  =  acosB -{- bcos  A.  J 

2c2  +  a2      2sin2C  +  sin2^ 


26.   Prove 


3abc  3  6  sin  A  sin  C 


27.  Multiplying  the  equations  in  (1)  of  example  25  by  —  a,  6,  and  c 
respectively,  and  adding,  we  obtain  the  law  of  cosines, 

62  +  c2-a2  =  26ccos^. 

28.  Prove  the  relations  in  (1)  of  example  25  directly  from  a  figure. 
Suggestion.     See  the  figures  in  §  52. 

61.   Area  of  a  triangle. 

(i)  In  terms  of  two  sides  and  the  sine  of  their  included  angle. 
Let  F  denote  the  area  of  any  triangle,  as  ^J5C  in  §  52. 
Then  F=^^  .i)C/2  =  be  sin  A/2.  [31] 

(ii)  In  terms  of  the  three  sides. 
By  [31]  and  [13],  we  have 

F=bcsm(A/2)GOs(A/2). 
Hence,  by  [28],  F  =  Vs(s-a)  (s-b)(s-c).  [32] 

(ill)  In  terms  of  one  side  and  the  sines  of  the  angles. 
In  [31],  putting  for  c  its  value  b  sin  C/sin  .B  (obtained  from 
the  law  of  sines),  we  have  ^ 

p^b^sinAsinC 

2  8inB  "-     -' 

EXERCISE  XXm 

1.  State  [31],  [32],  [33]  in  words. 

2.  Given  A  =  75°,  6  =  10,  c  =  40  ;  to  find  F. 

3.  Find  the  areas  of  the  triangles  in  examples  1-4  in  Exercise  XXI. . 

4.  Find  the  areas  of  the  triangles  in  examples  1-4  in  Exercise  XX. 

5.  Find  the  areas  of  the  triangles  in  examples  1-4  in  Exercise  XXII. 

6.  Find  the  areas  of  the  triangles  in  examples  1-4  in  Exercise  XIX. 


THREE  IMPORTANT  CIRCLES 


93 


62.  Circumscribed  circle.  Let  D  denote  the  diameter  of  the 
circle  circumscribed  about  the  triangle 

ABC.  ^       ~^\C 

Through  C  draw  the  diameter  COHj 
and  join  HB. 

Then  Z  A  =  Z.  H, 

Z  CBH  =  90°,  aY^ 7^ -r^B 

and  CH  =  D. 

.'.  sin  ^  =  sin  ^  =  a/D. 
.'.  a/sin  A  =  D.  (1) 

Compare  (1)  with  [23]. 

Example.  Find  the  radii  of  the  circles  circumscribed  about  the 
triangles  in  examples  1-4  in  Exercise  XIX. 

63.  Inscribed  circle.  Let  r  denote  the  radius  of  the  circle 
inscribed  in  the  triangle  ABC.  Join  the  center  0  with  the 
points  of  contact  Z>,  Ej  F.     Draw  OAj  OB,  OC. 

C 


FlO.  42 


Then  OD  =  OE  =  OF  =  r. 

A  ABC  =  A  BOC  -{-  A  COA  -\-AAOB. 
.-.  F=ar/2  -{-  br/2  +  cr/2 
=  r(a-\-b  -\-c)/2  =  rs. 


60 


94 


PLANE  TRIGONOMETRY 


Hence  r  ==  F/s, 

or  by  [32],  r  =  V(s  -a)(s-  b)  {s  -  g)/s.  (1) 

Compare  (1)  with  [29]. 

Example.  Find  the  radii  of  the  inscribed  circles  of  the  triangles  in 
examples  1-4  in  Exercise  XXII. 

64.  An  escribed  circle  of  a  triangle  is  a  circle  which  is 
tangent  to  one  side  of  the  triangle  and  to  each  of  the  other 
two  sides  produced. 

Ex.  1.  If  ra  denotes  the  radius  of  the  escribed  circle  tangent  to  the 
side  a,  prove  that 

r«  =  F/{s  -a)  =  Vs{s-b){s-c)/{s-a). 
A  ABC  =  AACD-\-/\  ABD  -  A  BCD. 
.:F=ra{b-\-c-a)/2 
=  ra{s-a). 


Fig.  44 


Ex.  2.     Find  the  radii  of  the  three  escribed  circles  of  the  triangles  in 
examples  1-3  in  Exercise  XXII. 


CHAPTER  VI 

RADIAN  MEASURE,  GENERAL  VALUES,  TRIGONOMETRIC 
EQUATIONS,  INVERSE  FUNCTIONS 

65.  A  radian  is  an  angle  which,  when  placed  at  the  center 
of  a  circle,  intercepts  between  its  sides  an  arc  equal  in  length 
to  the  radius  of  the  circle. 

That  is,  if  arc  AB  is  equal  to  the    ^        ^, ^^-^^ ' 

radius  OA,  then  ^^'  V  \^?- 

/  X  /      ^^ 

angle  AOB  =  a  radian.  /       \  /  \JU 

66.  Constant  value  of  the  radian.   [ .     \/  \  r  i^ 

In  fig.  45,  let  o  A 

Fig.  45 
arc  AB  =  OA  =  r  units. 

Then,  by  Geometry,  we  have  the  following  proportion : 
Z  AOB  :  180°  =  arc  AB :  semicircumference, 
i.e.  a  radian  :  180°  =  r :  irr. 

From  this  proportion,  we  have 

TT  radians  =  180°  =  2  right  angles.  [34] 

Formula  [34]  expresses  the  relation  between  radians,  degrees, 
and  right  angles,  and  should  be  fixed  in  mind. 

From[34],aradian=(180/7r)°,  or  (180/3.1416)°  (1) 

-57°.295779,  or  57°.3,  approximately  (2) 

=  57°  17'  44".8,  approximately.  (3) 

From  [34],  1°  =  (7r/180)  radian,  (4) 

or  a  right  angle  =  90°  =  (7r/2)  radians.  (5) 

Hence  270°  =  (3  tt  /  2)  radians, 

and  360°  =  2  tt  radians.  (6) 

95 


96  PLANE  TRIGONOMETRY 

Ex.  1.     Express  in  radians  the  angle  8°  15'. 

8°  15'  =  (33/4)°  =  (33/4)  (;r/180)  radian  by  (4) 

=  0. 144  radian,  approximately. 

Ex.  2.     Express  in  radians  the  sum  of  the  three  angles  of  a  triangle. 
Ex.  3.     Express  in  degrees  the  angle  5/8  radian. 

5/8  radian  =  (5/8)  (57°  17'  44".8)  by  (3) 

=  35°  48'  35".  5. 

67.  Radian  measure  of  angles.  In  fig.  45,  by  Geometry,  we 
have 

ZAOC  :  ZA0B  =  3,VG  ABC  :  arc  AB.  (1) 

Let  A  AOC  =  N  radians, 

arc  ABC  =  s  units, 
and  arc  AB  =  OA  =  r  units. 

Substituting  these  values  in  proportion  (1),  we  obtain 
N  radians  :  1  radian  =  sir. 

.-.  N  =  s/r.  [35] 

That  is,  the  number  of  radians  in  an  angle  at  the  center  of  a 
circle  is  equal  to  the  intercepted  arc  divided  by  the  radius. 

Cor.     If  r  =  l,    N  =  s. 

The  number  of  radians  in  an  angle  is  called  its  radian  (or 
circular)  measure. 

When  the  measure  of  an  angle  is  given  in  terms  of  tt  or  any 
other  numeral  or  numerals  and  no  angular  unit  is  expressed, 
the  radian  is  always  understood  as  the  unit. 

U.g.,  the  angle  2  is  an  angle  of  2  radians,  and  the  angle 
7r/2  is  an  angle  of  7r/2  radians. 

The  fraction  22/7  gives  the  correct  value  of  Tt  to  two  places  of  deci- 
mals, and  for  many  purposes  this  value  is  sufficiently  accurate. 

Ex.  1.  An  angle  at  the  center  of  a  circle  whose  radius  is  6  ft.  inter- 
cepts an  arc  of  3  ft.     Find  the  angle  in  degrees. 

The  angle  =  (3/5)  radian  §67 

=  (3  /  5)  (180  /  7t)°  =  34°^.  §  66 


RADIANS  97 

Ex.  2.  An  angle  of  42°  20'  is  at  the  center  of  a  circle  whose  radius  is 
10  ft.     Find  the  length  of  the  arc  intercepted  between  its  sides. 

Let  X  =  the  number  of  feet  in  the  intercepted  arc. 

Then        ic/10  =  the  number  of  radians  in  42°  20',  or  (127/3)° 

=  (127/3)  (;r/ 180).  by  [34] 

.-.  X  =  (127  /  64)  (22  /  7)  =  1j%%,  approximately. 
Hence  the  intercepted  arc  is  approximately  7^'^/^  ft.  long. 

Ex.  3.    From  [35]  find  s  in  terms  of  N  and  r,  and  r  in  terms  of  8 

and  JV^. 

EXERCISE  XXIV 

1.  Express  in  radians  two  positive  and  two  negative  angles  each  of 
which  is  coterminal  with  7t/4;  5;r/4;  3;r/2;  6;r/2;  7t/S;  2n:/S; 
7t/6;   bit/Q, 

Express  in  degrees  each  of  the  following  angles  : 

2.  2  7r/3.  3.    5;r/3.  4.    bit.  5.   3;r/4. 

Express  in  radians  each  of  the  following  angles  : 

6.  45°.  8.   90°.  10.   97°  25'.  12.   43°  25' 36". 

7.  135°.  9.   270°.  11.    175°  13'.  13.   38°  17'  23". 

Find  the  complement  and  the  supplement  of : 

14.   7r/4.  15.   2  7r/3.  16.   3;r/4.  17.   5;r/3. 

Find  the  trigonometric  ratios  of  each  of  the  following  angles  : 

18.    Tt/Q.  19.    ;r/4.  20.    ;f/3.  21.    ;r/2.  22.    it. 

23.  Two  angles  of  a  triangle  are  1/2  and  2/5.  Find  the  third  angle 
in  radians,  also  in  degrees. 

24.  The  difference  between  the  two  acute  angles  of  a  right-angled 
triangle  is  n/l.     Find  the  angles  in  radians,  also  in  degrees. 

25.  Express  in  radians,  also  in  degrees,  the  angle  subtended  at  the 
center  of  a  circle  by  an  arc  whose  length  is  15  ft.,  the  radius  of  the  circle 
being  20  ft.  i 

26.  The  diameter  of  a  graduated  circle  is  6  ft.,  and  the  graduations 
on  its  rim  are  5'  apart.  Find  the  distance  from  one  graduation  to  the 
next. 


98  PLANE  TRIGONOMETRY 

27.  Find  the  radius  of  a  globe  which  is  such  that  the  distance  between 
two  places  on  the  same  meridian  whose  latitudes  differ  by  1°  10'  may  be 
half  an  inch. 

28.  The  value  of  the  divisions  on  the  outer  rim  of  a  graduated  circle 
is  5'  and  the  distance  between  successive  graduations  is  0.1  in.  Find 
the  radius  of  the  circle. 

29.  Assuming  the  earth  to  be  a  sphere  and  the  distance  between  two 
points  1°  apart  to  be  69i  mi, ,  find  the  radius  of  the  earth. 

68.  Principal  values.  If  sin  6  =  —  \  12,  we  know  that  Q  is 
any  angle  which  is  coterminal  with  —  7r/6  or  —  57r/6.  Of 
this  series  of  values  the  smallest  in  size,  —  7r/6,  is  called  the 
principal  value  of  0. 

The  principal  value  of  an  angle  having  a  given  trigonometric 
i-atio  is  the  angle,  smallest  in  size,  which  has  this  ratio. 

Hence,  if  sin  6  or  esc  6  is  positive,  the  principal  value  of  6 
lies  between  0  and  7r/2 ;  if  sin  6  or  esc  6  is  negative,  the 
principal  value  of  6  lies  between  —  7r/2  and  0. 

If  tan  6  or  cot  6  is  -f,  the  principal  value  of  6  lies  between 
0  and  7r/2;  if  tan  6  or  cot  6  is  — ,  the  principal  value  of  6 
lies  between  —  7r/2  and  0. 

If  cos  6  or  sec  6  is  +,  the  principal  value  of  6  lies  between 
0  and  7r/2,  preference  being  given  to  the  positive  value;  if 
cos  B  or  sec  6  is  — ,  the  principal  value  of  B  lies  between  7r/2 
and  TT. 

69.  Angles  having  the  same  sine. 

Let  Z.  XOP  =  A,  A  XOP' =  ir  -  A. 

Then  sin  ZOP'  =:  sin  XOP,  esc  XOP'  =  esc  XOP, 

and  any  angle  which  has  the  same  sine  or  cosecant  as  A  is, 
P'  p        or  can  be  made,  coterminal  with  .1 

X.  y^         or  TT  —  A . 

^S^-*^  >/  In    this    chapter    n   will    denote 

M' -^^^^— j-i -L-x  any   positive   or   negative   integer, 

Pig.  46  including  zero. 


TRIGONOMETRIC  EQUATIONS  99 

When  n  is  even,  nir  +  A  includes  all  the  angles,  and  only 
those,  which  are  coterminal  with  A. 
Now,  when  n  is  even, 

nir  -[-  A  =  TiTT  +  (-  1)M.  (1) 

Again,  when  n  is  odd,  n  —  1  is  even,  and  (n  —  1)  ir  -{- (ir  —  A) 
includes  all  the  angles,  and  only  those,  which  are  coterminal 
with  IT  —  A. 

But  when  n  is  odd, 

{n-l)ir  +  {ir-A)=n'Tr  -A  =mr^-{-  1)M.         (2) 

From  (1)  and  (2)  it  follows  that  the  expression  nir  -\-  (—  1)"^ 
includes  all  the  angles,  and  only  those,  which  are  coterminal 
with  A  or  ir  ~  A. 

Hence  the  general  expression  mr4-(—  1)'^A  includes  all  the 
angles,  and  only  those^  which  have  the  same  sine  or  cosecant 
as  A. 

That  is,  the  general  value  of  0  in  the  equation 

sin  0  =  sin  A  (or  esc  ^  =  esc  ^)  is  nir  +  (—  1)"^^ 

E.g.,  if  sin  ^  =  sin(;r/3),  then  d  =  nTt  +  {-  l)«;r/3. 

Example.     Find  the  general  value  of  6,  if  sin  ^  =  —  1  /2.  (1) 

The  principal  value  of  d  in  (1)  is  —  7t/6. 

Substituting  for  —  1/2  in  (1)  its  identical  expression  sin(—  ;r/6),  we 
obtain  the  equivalent  equation 

sin^  =  sin(-;r/6).  (2) 

,.,0  =  n7r  +  {-1)-{-7t/6) 

=  ?i7r-(-l)»-7f/6.  (3) 

By  using  the  principal  value  of  d  in  (2),  we  obtain  the  simplest  expres- 
sion for  6  in  (3). 

Sometimes,  however,  in  solving  an  equation  it  is  advantageous  to  use 
some  other  than  the  principal  value  of  the  unknown  angle. 

Observe  that  the  unit  understood  with  6  and  A  is  the  radian.  If  the 
degree  were  the  unit,  we  would  write  d  =  n- 180°  +  (—  1)"^. 


100 


PLANE  TRIGONOMETRY 


Fig.  47 


70.  Angles  having  the  same  cosine. 
P  Let  Z  XOP  =  A,Z.  XOP'  =  -  A, 

Then  cos  XOP'  =  cos  XOP, 

,vf  and  sec  XOP'  =  sec  XOP, 

M  ' 

and  any  angle  which  has  the  same  cosine  or 

p*    secant  as  A  is,  or  can  be  made,  coterminal 

with  A  or  —  ^. 

Now  2  niT  ±  A    includes   all   the  angles,  and  only  those, 

which  are  coterminal  with  A  or  —A. 

Hence  2  rnr  ±  A  includes  all  the  angles,  and  only  those, 

which  have  the  same  cosine  or  secant  as  A. 

That  is,  the  general  value  of  6  in  the  equation 

cos  ^  =  cos  ^  {or  sec  6  =  sec  A)  is  2  nir  ±  A. 

E.g.,  if  cos  e  =  cos(;r/9),  then  e  =  2n7t  ±  7t/9,  or  2  n  •  180°  ±  20°. 
Example.     Find  the  general  value  of  0,  if  cos  ^  =  —  V3/2. 
Substituting  for  —  •v/3/2  its  identical  expression  cos  (5  ^/6),  we  obtain 
the  equivalent  equation 

cos^  =  cos(5;r/6). 
.-.  ^  =  2  n;r  ±  5  ;r/6,  or  2  n  •  180°  ±  150°. 

71.  Angles  having  the  same  tangent. 

Let  Z  XOP  =:A,Z  XOP'  =  tt -\-  A. 

Then  tan  XOP'  =  tan  XOP,  cot  XOP'  =  cot  XOP, 

and  any  angle  which  has  the  same  tangent  or  cotangent  as 
A  is,  or  can  be  made,  coterminal  with 
A  or  TT  +  ^. 

When  n  is  even,  nir  -\-  A  includes  M_ 
all  the  angles,  and  only  those,  which 
are  coterminal  with  A. 

When  n  is  odd,  7^  —  1  is  even,  and 
(n  —  l)TT  +  {ir-\-  A),  i.e.  nir  -\-  A,  includes  all  the  angles,  and 
only  those,  which  are  coterminal  with  ir  -\-  A. 


Xl 

P 

m' 

'ry^ 

^0   • 

M  ' 

V 

Fig.  48 

TRIGONOMETRIC  EQUATIONS  101 

Hence  nir  -j-  A  includes  all  the  angles^  and  only  those,  which 
have  the  same  tangent  or  cotangent  as  A. 

That  is,  the  general  value  of  0  in  the  equation 

tan  B  =  tan  A  (or  cot  $  =  cot  A)  is  nir  +  A. 

E.g.^  if  cote  =  cot(-  ;r/5),  6  =  nit  -{-  {-  it/b),  or  mt  —  7t/6. 

Example.     Find  the  general  value  of  d,  if  tan  6  =  —  y/S. 
Substituting  for  —  V3  its  identical  expression  tan(—  ;f /3),  we  obtain 
the  equivalent  equation 

tane  =  tan(— ;r/3). 
.-.  e  =  n7t  -  7t/S,  or  n  •  180°  -  60°. 

72.  A  trigonometric  equation  is  an  equation  which  involves 
one  or  more  trigonometric  ratios  of  one  or  more  unknown 
angles,  as  tan  ^  =  1  or  2  sin^  $  +  V^  cos  ^4-1  =  0. 

To  solve  a  trigonometric  equation  is  to  obtain  an  expression 
for  all  the  angles  which  will  satisfy  it. 

The  first  step  in  solving  a  trigonometric  equation  is  to  solve 
it  algebraically  for  some  trigonometric  ratio  of  the  unknown 
angle ;  the  second  step  is  to  apply  one  or  more  of  the  princi- 
ples in  §§  69-71. 

Some  elementary  types  of  trigonometric  equations  are  solved 
below. 

Ex.  1.     Solve  the  equation  sin2e  =1/4. 
First  step.  sin  ^  =  ±  1  /2. 

Second  step.  .-.  sin  6  =  sin  ( ±  tt /6). 

.■.d  =  n7t-h{-l)»{±7t/6)  =  n7f±7e/6.     §69 

Ex.  2.     Solve  the  equation  cos^  ^  =  1  /2. 

Denote  the  two  values  of  cos  d  by  cos  di  and  cos  02. 

Then  cos  ^i  =  V2/2  =  cos  (7r/4),  (1) 

and  cos  ^2  =  -  V2  /2  =  cos  (5  it/ 4).  (2) 

From  (1),  ^1  =  2  ihtc  ±  7r/4.  (3) 

From  (2),  ^2  =  2  n2?r  ±  5  7r/4  =  (2  W2  ±  1)  ;r  ±  ;f /4.  (4) 


102  PLANE  TRIGONOMETRY 

When  n  is  even,  mt  ±7t  /^  includes  2  n\7t  db  7^/4  in  (3) ;  and  when  n 
is  odd,  nit  ±Tt /^  includes  (2  n2  ±  1)  tt  ±  7r/4  in  (4). 

Hence  Q  =  mt±,  it  /^. 

Observe  that,  by  using  5  7r/4  instead  of  the  principal  value  of  d^  in  (2), 
the  two  expressions  for  6  in  (3)  and  (4)  are  more  readily  combined  into   \ 
one  expression. 

Ex.  3.     Solve  the  equation  2  sin2  ^  -f  V3  cos  ^  +  1  =  0.  (1) 

Putting  for  sin^  d  its  identical   expression  1  —  cos^  6^  we  obtain  the 
equivalent  equation 

2  cos2  6*  -  V3  cos  ^  -  3  =  0,  (2) 

which  involves  only  one  function  of  the  unknown  angle  6. 

Factor,  (cos  6  -  V3)  (2  cos  6  +  V3)  =  0.  (3) 

By  Algebra,  (3)  is  equivalent  to  the  two  equations 

cos^=:V3,  cos^  =  -V3/2. 

Now,  by  §  26,  cos  ^  =  V^  is  impossible, 

and  cos^  =  - V3/2  =  cos(5;r/6) 

gives  e  =  2mt  ±bTt/Q.  §70 

Ex.  4.     Solve  the  equation  tan  bd  =  cot  2  6. 

Substituting  for  cot  2  0  its  identical  expression  tan(7r/2  —  2  0),  we 
obtain  the  equivalent  equation 

tan50  =  tan(7r/2  -2  0). 
.-.  50  =  n7r  +  (;r/2 -2  0).  §71 

.-.  0  =  (w;r  +  ;r/2)/7. 

Ex.  5.     Solve  tan  0  sec  0  =  -  y/2.  (1) 

Putting  for  sec0  its  identical  expression  Vtan^  0  +  1,  we  obtain  the 
equivalent  equation 

tan  0  Vtan2  0+1  =  -  V2.  (2) 

Square,  tan*  0  +  tan2  0  =  2. 

Factor,  (tan2  0  +  2)  (tan2  0  -  1)  =  0. 

Hence  tan  0  =  ±  V-  2  or  ±  1.  (3) 

Now  tan  0  =  ±  V—  2*  is  impossible. 

Since  tan  0  sec  0  is  negative  in  (1),  tan  0  and  sec  0  are  opposite  in 
quality,  whence  0  is  in  the  third  or  fourth  quadrant. 

Hence  tan  0  =  +  1    gives    0  =  2ii7r  +  5;r/4, 

and  tan  0  =:  -  1     gives    0  =  2  n7r  —  7r/4. 


TRIGONOMETRIC  EQUATIONS  103 

Observe  that  the  sohitions  of  (3)  which  do  not  satisfy  (1)  were  intro- 
duced by  squaring  (2). 

Ex.  6.  Given  sin  ^  =  —  1/2  and  tan  ^  =  1  /  V3 ;  to  find  the  general 
value  of  6. 

Since  sin  ^  is  — ,  and  tan  0  is  + ,  0  must  be  in  the  third  quadrant. 

The  smallest  positive  angle  in  the  third  quadrant  which  will  satisfy 
sin  ^  =  —  1/2  is  7  7r/6,  and  this  angle  satisfies  also  tan  6  =  l/y/S. 

Hence  d  =  2n7C  +  7  tc /6,  or  2n-  180°  +  210°. 

EXERCISE  XXV 

Solve  each  of  the  following  equations : 

1.  sin2(9  =  l.  3.    tan2^=l.  5.    cos2^=l/4. 

2.  csc-2  6  =  2.  4.    cot2  ^  =  3.  6.    sec2  ^  =  4/3. 

7.  2  sin2  ^  +  3  cos  ^  =  0.  10.    sin^  ^  -  2  cos  0  4-  1  /4  =  0. 

8.  cos2  ^  -  sin  0  =  1  /4.  11.    sin  ^  +  cos  ^  =  V2. 

9.  2  V3  cos2  d  =  sin  0.  12.   4  secS  d-1  tan2  0  =  3. 

13.  tan  0  +  cot  0  =  2. 

14.  tan2  0  -  (1  +  V3)  tan  0  +  V3  =  0. 

15.  cot2  0  +  (V3  + l/V3)cot<?  + 1  =0. 

16.  tan2  d  4-  cot2  d  =  2. 

17.  tan  0  +  sec  0  =  3.  22.   sin  2  0  =  cos  3  0. 

18.  2  sin  0  =  1  +  cos  0.  23.    cos  m0  =  sin  r  6. 

19.  sin50  =  l/V2.      ■  24.    sin^cos  0  =  1/2. 

20.  cos  5  0  =  cos  4  0.  25.    sin  0  cos  0  =  -  V3 /4. 

21.  cot  0  =  ianr  0.  26.    sec  0  esc  0  =  -  2. 

27.  tan  2  0  tan  0=1. 

Find  the  most  general  value  of  6  that  satisfies  : 

28.  cos  0  =  -  1  /  V2  and  tan  0  =  1. 

29.  cot  0  =  -  V3  and  esc  0  =  -  2. 

30.  If  cos(^  -  B)  =  1/2  and  sin  {A  +  B)  =  l/2,  find  the  smallest 
positive  values  of  A  and  B  and  also  their  general  values. 


104  PLANE  TRIGONOMETRY 

31.  If  tan (^  -  :B)  =  1  and  sec {A-\-  B)  =  2/  V3,  find  the  smallest 
positive  values  of  A  and  B  and  also  their  general  values. 

73.  The  addition  and  subtraction  formulas  and  those  for 
the  sum  and  difference  of  sines  or  cosines  are  often  useful  in 
solving  certain  types  of  trigonometric  equations. 

E.g.,  take  the  equation  a  sin  0  +  b  cos  9  =  c.  (1) 

Let  A  denote  the  principal  value  of  tan  -i  (a/6) ;  then  tan  ^  =  a/6, 
and  from  the  fundamental  relations  in  §  24  we  obtain 


sm 


^  =  a/Va2  +  62,  cos^  =  6/Va-^  +  62.  (2) 


Dividing  both  members  of  (1)  by  va2  +  6'2  and  substituting  for  the 
resulting  coefficients  of  sin  e  and  cos  6  their  values  given  in  (2),  we  have 


sin  A  sin  6  -\-  cos  Acosd  =  c/  Va2  +  62.  (3) 


Let    B    denote    the    principal    value    of    cos-i  (c/ Va2  +  62) ;    then 
cos^  =  c/ Va2  +  62,  and  by  [10]  from  (3)  we  obtain 

cos  {e  —  A)  =  cos  B. 

.'.e-A=2n7t±B.  §70 

.:  e  =  2n7t  +  A  ±  B,  OT  2  n  ■  ISO  -\-  A  ±  B.  (4) 

In  (4)  we  have  the  solution  of  any  equation  of  the  type  (1)  when  a 
and  6  are  real  and  c  <  or  =  Va2  +  62,  arithmetically. 

Ex.  1.     Solve  4  sin  ^  +  3  cos  0  =  5,  given  tan  53°  7'  45''  =  4/3. 
Here  A  =  tan-  ^  (4  /  3)  =  53°  T  45", 

and  B  =  cos-i  (5  /  V42  +  32)  =  cos-i  1=0. 

.'.d  =  2n- 180°  +  53°  7'  45". 

Ex.2.     Solve  sin  ^  +  sin  5  0  =  sin  3  e.  (1) 

By  [19],  sin  ^  +  sin  5  ^  =  2  sin  3  ^  cos  2  6.  (2) 

From  (1)  by  (2),   2  sin3  5cos2  ^  =  sin  3  ^. 

.-.  sin3^(2cos2^- 1)  =  0. 
From  sin 3  ^  =  0,  Sd  =  mt. 

From  cos2^  =  1/2,  2  ^  =  2  w;r  ±  ;r/3. 

Hence  6  =  nit/Z  or  mt  ±  ^/6. 


TRIGONOMETRIC  EQUATIONS  105 

EXERCISE  XXVI 

Solve  each  of  the  following  equations  : 

1.  sin  ^  +  sin  7  ^  =  sin  4  d.  7.  sin  ^  +  sin  3  0  +  sin  5  ^  =  0. 

2.  cos^  +  cos3^  =  cos2^.  8.  cos  ^  +  cos2^  +  cos3  ^  =  0. 

3.  sin  ^  —  sin  3  ^  =  cos  2  d.  9.  V3  cos  6  -\-  sind  =  V2. 

4.  cos  ^  +  cos  7  ^  =  cos  4  6.  10.  sin  ^  +  cos  ^  =  V2. 

-6.  sin  4  ^  —  sin  2  ^  =  cos  3  6.  11.    V^  sin  ^  -  cos  0  =  V2. 

6.  sin  7  ^  =  sin  ^  +  sin  3  6.  12.    sin  ^  +  cos  ^  =  V2  cos  {7C/b). 

13.  5  sin  ^  +  2  cos  ^  =  V29,  given  tan  68°  12'  =  5/2. 

14.  2  sin  ^  -  3  cos  ^  =  V13/2,  given  tan  33°  41'  24''  =  2/3. 

74.  Trigonometric  functions.  A  quantity  whose  value  depends 
upon  one  or  more  other  quantities  is  called  a  function  of  these 
quantities. 

E.g.,  the  circumference  or  the  area  of  a  circle  is  a  function  of  the 
radius ;  the  area  of  a  rectangle  is  a  function  of  the  base  and  the  altitude  ; 
the  volume  of  a  rectangular  parallelepiped  is  a  function  of  the  three 
dimensions. 

Since  the  trigonometric  ratios  of  an  angle  depend  upon  the 
size  of  the  angle,  they  are  often  called  the  trigonometric  func- 
tions of  the  angle. 

Beside  the  six  trigonometric  functions  defined  in  §  21,  there 
are  two  others  which  are  frequently  used  : 

1  —  cos  A  is  called  the  versed  sine  of  A,  written  vers  A. 
1  —  sin  ^  is  called  the  coversed  sine  of  A,  written  covers  A. 
As  a  ratio,  vers  A  =  (OP  —  OM)/OPj 

and  covers  A  =  (OP  —  MP)  /  OP. 

Note.  After  the  analogy  of  vers  A  and  covers  A  many  other  trigo- 
nometric ratios,  or  functions,  of  A  might  be  invented  and  named. 

E.g.,  1  +  cos  ^,  or  {OP  +  OM)  /  OP,  is  sometimes  called  the  suversed 
sine  of  A,  and  sec  J.  —  1,  or  (OP  -  OM)  /  OM,  the  external  secant  of  A. 


106  PLANE  TRIGONOMETKY 

Observe  that  the  trigonometric  lines  defined  in  §  32  are  also 
functions  of  the  angle  or  of  its  measuring  arc. 

But  by  the  trigonometric  functions  we  usually  mean  the 
trigonometric  ratios. 

75.    Inverse  trigonometric  functions. 

By  §  23,  if         G  =  sin  0,   then  6  =  sin-^c.  (1) 

As  the  value  of  sin  6  depends  on  the  value  of  the  angle  B, 
so  the  value  of  sin~^c  depends  on  the  value  of  c. 

Hence,  on  the  one  hand,  the  sine  is  a  function  of  the  angle, 
and,  on  the  other  hand,  the  angle  is  a  function  of  the  sine. 

An  angle  is  often  called  the  inverse  function  of  any  one  of 
its  trigonometric  ratios.  Thus  sin~^c  is,  often  read  inverse 
sine  c  ;  cos~^<?,  inverse  cosine  c  ;  tan~^c,  inverse  tangent  c,  etc. 

Note.  If  A  denotes  all  the  angles  which  have  the  same  sine,  then 
sin-i  (sin  A)^A  and  sin  (sin-i  c)  ^  c  ;  hence,  if  we  regard  sin  as  denot- 
ing the  operation  of  finding  the  sine  of  and  sin-^  as  denoting  the  operation 
of  finding  the  angle  whose  sine  is,  then  sin—^  and  sin  denote  inverse 
operations,  i.e.  operations  each  of  which  undoes  what  the  other  does. 
For  this  reason  sin- 1  c  is  called  the  inverse  sine  of  c. 

Sin— ic  is  read  also  anti-sine  c,  or  arc  sinec. 

For  each  value  of  0,  sin  0  has  a  single  definite  value. 

For  each  value  of  c,  sin"~^  c  has,  by  §  23,  an  indefinite  number 
of  values. 

Thus  the  trigonometric  functions  are  single-valued,  while  the 
inverse  trigono7netric  functions  are  multiple-valued. 

E.g..,  if  tan  6>  =  1,  6  =  tan-i  \  =  mt  -\-  it / A.,  and  the  principal  value  of 
tau-il  is  ;r/4. 

If  cos  ^  =  1  /2,  e  =  cos-i  (1  /2)  =  2  n;r  ±  7r/3,  and  the  principal  value 
of  cos-i(l/2)  is  Tt/Z. 

If  sin^  =  V2/2,  ^  =  sin-i(V2/2)  =  htt  +  (-  l)«;r/4,  and  the  prin- 
cipal value  of  sin-i(  V2/2)  is  ;r/4. 

Example.  If  a  and  c  are  positive  and  c  <  1,  and  the  inverse  functions 
are  restricted  to  their  principal  values,  show  that 

sin-^c  +  cos— ^c  =  7r/2,  tan— 'a  +  cot— ^a  =  ;r/2. 


INVERSE  FUNCTIONS  107 

76.    Sum  and  difference  of  two  inverse  tangents. 
To  prove  tan"^  m  ±  tan~^  n  =  tan~^ [36] 

Let  A  =  tan~^  m,  B  =  tan~^  n.  (1) 

Then  tan  A  =  m,  tan  B  =  n.  (2) 

Using  the  notation  in  (1),  [36]  becomes 

A  ±B  =  tan-^  T ,  or  tan  (A  ±  B)  = (3) 

1  q:  77171  ^       1  qi  mn         ^  ^ 

To  prove  (3),  by  §  37  we  have 

tan  A  ±  tan  B 


tan  {A  ±  B) 


1  qi  tan  A  tan  B 

(m  ±  n)  /  (1  ip  mn).  by  (2) 


The  proof  above  ilhistrates   the  method  of  procedure  in 
establishing  relations  between  inverse  functions. 

Ex.1.     Prove  that  sin- 1(3/5) -f  sin- 1(8/ 17)  =  sin- 1(77/ 85).  (1) 

Let                ^^sin-i(3/5),                             5  =  sin-i(8/17).  (2) 

Then       sin  ^  =  3/5,                                   sin  5  =  8/17.  (3) 

.-.cos  ^  =  4/5,                                   cos  B=  15/17.  (4) 
Using  the  notation  in  (2),  (1)  becomes 

^  +  1?  =  sin- 1(77/ 85),  or  sin  (^  +  J?)  =  77/85.  (6) 
To  prove  (5),  by  [7]  we  have 

sin  {A  -\-  B)  =  sin  ^  cos  J5  +  cos  A  sin  B 

Ex.  2.     Prove  that  cos-i  (4/5)  +  tan- 1  (3/5)  =  tan- 1  (27/11).  (1) 

Let                ^  =  cos-i(4/5),                             B  =  tan-i(3/5).  (2) 

Then       cos  ^  =  4 / 5,      tan  A  =  S/4,       tan  B  =  3/5.  (3) 
Using  the  notation  in  (2),  (1)  becomes 

^  +  B  =  tan- 1(27/11),  or  tan  (^  +  5)  =  27/11.  (4) 


108  PLANE  TRIGONOMETRY 

To  prove  (4),  by  [11]  and  (3)  we  have 

3/4  +  3/5      _27 


tan  {A  +  B) 


1- (3/4)  (3/5)      11 


Ex.  3.     Prove  that  2  tan-i  (1/3)  +  tan-i  (1  /7)  =  7r/4.  (1) 

Let  ^  =  tan-i(l/3),  5  =  tan-i(l/7).         (2) 

Then       tanJ.  =  l/3,  tanJ5=l/7.  (3) 

Using  the  notation  in  (2),  (1)  becomes 

2A  +  B=7t/4,         or         tan(2^  +  5)  =  tan(7r/4)  =  1.     (4) 

^     r.KH  r.  .         2tan^  2/3         3 

By  [151,  tan  2  ^  = =  —^ =  -  •  (6) 

^  ^     -"  l-tan2A      1-1/9      4  ^  ' 

To  prove  (4),  by  [11],  (3),  and  (5)  we  have 

3/4  +  1/7      _, 


tan  {2A-J-  B) 


1- (3/4)  (1/7) 


Ex.  4.     Solve  the  equation  tan-i  2  x  +  tan-i  3  ic  =  ;r/4.  (1) 

2  ic  +  3  ic 
By  §76,     tan-i2x  +  tan-i3x  =  tan-i ^^^-  (2) 

From  (1)  by  (2),  tan-i^^±?|  =  ^  (3) 

1  —  D  X-^  4 

-r,  /ex  2X+SX  ^  7t  ^ 

.-.  x  =  1/6  or  -  1. 
x  =  l/6  satisfies  (1)  for  the  principal  values  of  tan-i 2 x  and  tan— ^ 3 x. 
X  =  —  1  satisfies  (1)  for  the  values 

tan-i(-2)  =  116°33'55^ 

tan-i(-3)  =  -71°33'55'^ 

Ex.  6.    Solve  the  equation  tan- 1 +  tan- 1 =  tan-  ^  -  7 ).  ( 1 ) 

X  —  1  X 

tan-i h  tan-i =  tan- ^ (2) 

X  - 1  X  1  —X 

From  (1)  by  (2),      ^^^-^  +  '^  =  _  7^  or  x  =  2. 
1  —  X 

X  =  2  satisfies  (1)  when  for  tan-i  (-  7)  we  take  98°  T  48". 


INVERSE  FUNCTIONS  109 


EXERCISE   XXVn 

1.  Find  the  value  of  vers  (tt  /  6) ,  vers  (tt  /  4) ,  vers  (tt  / 3) ,  vers  (3  tt  / 4), 
versO,  vers(;r/2),  vers  7f ,  vers  (3^/2). 

2.  Find    the   value    of    covers  (tt/G),    covers  (;r/4),    covers  (;r/3), 
covers (3  TT/ 4),  covers  0,  covers (;r/2),  covers  ;r,  covers (3  ;r/ 2). 

3.  Express  in  radians  the  general  value  of 

sin-i(V2/2);  sin-i(- V3/2);  cos-i(V3/2);  cos-i(-l/2); 
tan-i(V3/3);  tan-i(-V3);  cot-i(-l);  cot-i(V3/3). 

Prove  each  of  the  following  relations  for  principal  values  : 

4.  tan-12  +  tan-i0.5  =  7r/2. 

5.  tan-i  7  +  tan-i  3  =  153°  26'  6'^5,  given  0.5  =  tan  26°  33'  53".5. 

6.  tan-i tan-i — ; —  =  -• 

n  m  +  n      4 

7.  tan-i(l/7)  +  tan-i(l/13)  =  tan-i(2/9). 

8.  2tan-i(2/3)  =  tan-i(2/3)  +  tan-i(2/3)  =  tan-i(12/6). 

9.  tan-i(3/4)  +  tan-i(3/5)  -  tan-i(8/19)  =  7r/4. 

Add  tan- 1(3/ 5)  to  tan- 1  (3/ 4)  and  then  subtract  tan- 1  (8/19). 

10.  sin-i(3/5)  +  sin-i(8/17)  =  sin-i(77/85). 

11.  cos-i(4/5)  +  cos-i (12/13)  =  cos-i (33/65). 

12.  cos-i(4/5)  +  tan-i(3/5)  =  tan-i(27/ll). 
Solve  each  of  the  following  equations  : 

13.  tan-ix  +  tan-i  {1  -  x)  =  tan-i  (4/3). 

14.  tan-.^-t  +  Un-i^±i  =  |. 

x-2  x+2      4 

15.  tan-ix  +  2cot-ijc  =  2;r/3. 

16.  tan-i  (X  +  1)  +  tan-i  (x  -  1)  =  tan- 1  (8/31). 

17.  cos-i h  tan-i  — =  — -  • 

x2  +  1  x2  -  1        3 

18.  sin-ix  +  sin-i2x  =  7r/3. 

19.  sin-i  (5/x)  +  sin-i  (12 /x)  =  7t/2. 

Suggestion.     Observe  that  sin- 1  (5/x)  and  sin-i(12/x)  are  comple- 
mentary angles. 


CHAPTER   VII 

PERIODS,    GRAPHS,    IMPORTANT    LIMITS,    COMPUTATION    OF 
TABLE,   HYPERBOLIC  FUNCTIONS 


77.  Periods  of  the  trigonometric  functions.  As  an  angle 
increases  from  0  to  2  7r  radians,  its  sine  first  increases  from 
0  to  1,  then  decreases  from  1  to  —  1,  and  finally  increases 
from  —  1  to  0.  As  the  angle  increases  from  2  tt  to  4  tt  radians, 
the  sine  again  goes  through  this  same  series  of  changes.  Thus 
the  sine  goes  through  all  its  changes  while  the  angle  increases 
by  2  7r  radians.  This  is  expressed  by  saying  that  the  period 
of  the  sine  is  2  it. 

Similarly  the  cosine,  secant,. or  cosecant  goes  through  all 
its  changes  while  the  angle  increases  by  2  tt  radians. 

The  tangent  or  cotangent,  however,  goes  through  all  its 
changes  while  the  angle  increases  from  0  to  tt  radians. 

Hence  the  period  of  the  sine,  cosine,  secayit,  or  cosecant  is 
217  radians;    while  the  period  of  the  tamjent  or 
cotangent  is  ir  radians. 

Since,  as  the  angle  increases,  each  trigonometric 
function  goes  through  again  and  again  the  same 
series  of  values,  these  functions  are  called  periodic 
functions. 


78.    Curve  of  sines.    Let  OX  =  OP^^^  1, 

^3 

// 

k> 

ZXOPi  =  7r/6, 

^ 

\ 

and                                ZZOP2  =  7r/3. 

0 

M,M, 

Then,  if                                ^  =  Z  XOP, 

Fig.  49 

we  have,  when 

^=0,  7r/6,     7r/3,    7r/2,  27r/3,  57r/6,  tt, 

77r/6, 

47 

r/3 

J 

Sin^=0,  iV/iPi,  M^P^,  OPs,  MoTo,  M^P^,  0, 

110 


il/iPi,  -il/aA, 


CURVE  OF  SINES 


111 


The  series  of  values  through  which  sin  0  passes  as  the  angle  0 
increases  can  be  graphically  represented  by  means  of  the  points 
of  a  curve  constructed  as  follows  : 

Let  OX  and  0  F  in  fig.  50  be  two  fixed  straight  lines  at  right 
angles  to  each  other,  and  let  an  angle  of  one  radian  be  repre- 
sented by  a  unit  of  length  along  OX.  Take  OR  =  ir  units, 
i.e.  about  3}  units  of  length ;  also  take  BX  =  ir  units,  and 
OZ'  =  —  TT  units. 

Then  OR  represents  tt  radians ;  OX,  2  tt  radians ;  and  OX', 
—  TT  radians. 


Fig.  50 


Lay  off  on  OX  the  distances  representing  0,  7r/6,  tt/S,  it  12, 
2  7r/3,  etc.  At  N,  the  end  of  7r/6,  draw  NP  X  OX  and  equal 
to  MxP\  in  fig.  49 ;  at  the  end  of  7r/3  draw  a  perpendicular 
equal  to  M^P^ ;  and  so  on.  Through  the  ends  P,  P',  P",  •of 
these  perpendiculars  draw  a  smooth  curve. 

Then  if  from  any  point  in  this  curve,  as  P,  we  draw  PN  ±  OX, 
the  directed  line  NP  represents  the  sine  of  the  angle  whose 
radian  measure  is  represented  by  the  directed  line  ON. 

This  curve,  called  the  curve  of  sines,  or  sinusoid,  consists 
of  portions  similar  to  OP"RQX  placed  one  after  another. 

This  illustrates  graphically  that  the  period  of  the  sine  is 

2  IT. 


79.    Curve  of  cosines.     Using  fig.  49,  we  obtain,  when 
e=0,    7r/6,  7r/3,   7r/2,  27r/3,  57r/6,  tt,   77r/6,  47r/3, 


cos^=+l,  OMi,  OM^,  0, 


01/2,  -OMi,  -1,  -OMi,  -OM^, 


112 


PLANE  TRIGONOMETRY 


Lay  off  on  OX  the  distances  representing  0,  tt/G,  7r/3,  7r/2, 
2  7r/3,  etc.  At  the  points  thus  determined  erect  perpendiculars 
equal  to  OX,  OM^,  OM^,  0,  —  OM^,  etc.,  and  trace  a  curve  through 
the  points  P,  P',  P",  etc. 

The  curve  thus  obtained  is  called  the  curve  of  cosines. 


P  Y 


Fig.  51 


Observe  that  the  curve  of  cosines  has  the  same  form  as  the 
curve  of  sines.  These  two  curves  differ  only  in  their  position 
with  reference  to  the  origin  0. 

80.  Curve  of  tangents.  Using  fig.  49,  where  XOP  =  27r/5, 
we  obtain,  when 

0=0,  7r/6,  7r/3,  27r/5,  7r/2,  -7r/6,  -7r/3,  -27r/5,  -7r/2,.  • ., 
tan  ^=0,  X^i,  XT^,  XT,      oo,       -XT^,  -XT^,  -XT,      -oo,     •    •. 

Proceeding  as  in  §  78,  we  obtain  the  curve  represented  by  the 
continuous  lines  in  fig.  62.  This  graph  is  called  the  curve  of 
tangents. 

Since  tan(7r/2)  =  oo,  the  curve  of  tangents  will  have  no 
point  on  RK,  but  the  infinite  branch  P'P"  will  approach 
indefinitely  near  to  the  line  RK  without  ever  touching  it. 
The  same  is  true  of  the  infinite  branch  PiPz  with  reference 
to  the  line  RiKi. 

The  curve  of  tangents  will  evidently  consist  of  an  unlimited 
number  of  similar  but  disconnected  branches,  any  one  of 
which  is  parallel  to  the  branch  P^OP". 


CURVE  OF  TANGENTS 


113 


Curve  of  cotangents.     Using  fig.  49,  by  §  30,  we  obtain,  when 

^=0,  7r/10,  7r/6,  7r/3,  7r/2,  27r/3,  Bir/d,   Qtt/IO,  tt,    .••, 
cot^=oo,  XT,      XT^,  ZTi,  0,        -XT^,  -XT^,  -XT,     -oo,.--. 


.''liT 


Fig.  52 


Proceeding  as  above,  we  obtain  the  curve  represented  by 
the  dotted  lines  in  fig.  52  as  three  of  the  infinite  number  of 
disconnected  branches  of  the  curve  of  cotangents. 

81.   Curve  of  secants.     Using  fig.  49,  we  obtain,  when 

^=0,    7r/6,  7r/3,  27r/5,  7r/2,  -7r/6,  -7r/3,  -27r/5,  -7r/2,.  •  •, 
sec^=+l,  OTi,  07^2,  OT,       oo. 


0^1,   OTs,    or. 


00, 


Proceeding  as  above,  we  obtain  BPK  and  B'P'K'  as  two 
branches  of  the  curve  of  secants;  repetitions  of  these  two 
branches  make  up  the  rest  of  the  curve. 


114 


PLANE  TRIGONOMETRY 


Similarly  the  curve  of  cosecants  can  be  constructed,  three 
branches  of  which  are  represented  by  the  dotted  curved  lines 
in  fig.  53. 


2T 


Fig.  53 


82.  *Lt(sin  0/6)  =  lt(tane/e)  =  1,  when  6  =  0,  the  unit  of 
6  being  the  radian. 

Let  0  be  the  radian  measure  of  any  positive  acute  angle 
A  OP.     Draw  the  arc  AP,  the  chord  AP,  PM  ±  OA,  A  r±  OA. 

*When,  according  to  its  law  of  change,  a  variable  will  become  and  remain 
less  in  size  than  any  assignable  constant  value  but  will  never  become  zero,  the 
variable  is  called  an  infinitesimal,  or  is  said  to  approach  zero  as  its  limit. 

When  a  variable  approaches  a  constant  so  that  their  difference  is  an  infini- 
tesimal, the  variable  is  said  to  approach  the  constant  as  its  limit,  or  the 
constant  is  said  to  be  the  limit  of  the  variable. 

Lt  (sin  e  /9)  is  read  the  limit  of  sine  /e;  0  =  0  is  read  e  approaches  0  as  its 
limit  or  e  is  an  infinitesimal. 

The  reciprocal  of  an  infinitesimal  is  an  infinite  and  is  denoted  by  co. 

Any  number  which  is  neither  an  infinitesimal  nor  an  infinite  is  called  n  finite 
number 


LIMIT  OF  SIN  0/0  115 

Then  AOAP  <  sector  0.4P  <  A  OA  T.  (1) 

By  Geometry,  the  area  of  the  sector  OAP  is  OA  -arc  AP/2. 
Hence,  from  (1),  we  have 

OA  •  MP  <  OA  .  arc  .4P  <  OA  -  A  T.  (2) 

Dividing  each  member  of  (2)  by  OA^,  we  obtain 
MP  I  OP  <2iVGAP/0A<AT/0A; 
that  is,  sin  ^  <  ^  <  tan  0.  (3) 

Dividing  the  members  of  (3)  first  by  sin  0  and  then  by  tan  0, 
we  have 

1  <  ^/sin  0  <  sec  (9,    (4)  P, 

and  cos  0  <  ^/tan  ^  <  1. 

Let  0  =  0. 

Then  sec  ^  =  1,  ^^^ 

and  cos  0  =  1. 

Hence  the  ratio  0/s'mO  or  the  ratio  0/tsinO  lies  between 
1  and  a  variable  whose  limit  is  1 ;  hence  the  limit  of  each  of 
these  ratios  is  1. 

Now  _^/sin(-^)    = -0/(-smO)  =  0/smO. 

Hence  It  [-  ^/sin(-^)]  =  lt(^/sin^)        =  1,  when  ^=.0. 
Also,     lt[-  (9/tan(-  ^)]  =  It  (O/tan  0)       =  1. 

Hence  the  theorem  holds  true  whether  0  is  positive  or 
negative. 

Example.     To  find  sin  V  and  cos  1'. 

By  §  66,  r  =  (0.00  029  088  821  + )  radian.  (!') 

By  (3)  and  (1'),      sin  V  <  V  in  radians  <  0.00  029  088  822.  (2') 

Again,  cos  V  =  Vl  -  sin^  V  >  Vl  -  (.0003)2  >  .99  999  99^ 

Hence,  to  seven  places, 

cos  1'  =  0.99  999  99  +  .  (3') 

By  (4),  sin  ^  >  ^  cos  d. 

.'.  sin  1'  >  (1'  in  radians)  cos  V. 


116  PLANE  TRIGONOMETRY 

.-.  sin  r  >  0.00  029  088  821  x  0.99  999  99  by  (1'),  (30 

>  0.00  029  088  821  x  (1  -  0.00  000  01), 
or  sin  1'  >  0. 00  029  088  820 + .  (4^ 

From  (2')  and  (4')  it  follows  that  to  ten  places 

sin  r  =  0.00  029  088  82  +  ,  (5') 

the  eleventh  figure  being  0  or  1. 

This  example  affords  a  good  illustration  of  the  use  of  corollary  1,  below. 

Two  very  important  corollaries  to  the  theorem  above  are  the 
following : 

Cor.  1.  If  Q  is  very  small,  it  can  be  substituted  for  either 
sin  6  or  tan  6  in  approximate  calculations  ;  or  vice  versa. 

Cor.  2.  i/*  6  is  an  infinitesimal  and  it  is  substituted  for  either 
sin  6  or  tan  9  in  any  function,  the  limit  of  the  function  will  not 
be  changed. 

83.    Convenient  formulas  for  computing  the  sine  and  cosine  of 

any  angle  are 

<f>2       W,*        d,«       d>8        </,io 

cos<A-l-|  +  |-|  +  |-^  +  ...,        .     (c) 

where  (f>  is  the  radian  measure  of  the  angle. 

For  the  proof  of  (b)  and  (c),  see  §  94  in  Taylor's  Calculus. 

The  sine  of  any  angle  is  arithmetically  equal  to  the  sine  or 
cosine  of  some  angle  not  greater  than  7r/4  ;  hence  its  value  can 
be  computed  by  taking  <^  in  (b)  or  (c)  not  greater  than  7r/4. 

Likewise  the  cosine  of  any  angle  can  be  computed  by  taking 
<^  in  (c)  or  (b)  not  greater  than  7r/4. 

Observe  that  for  <^  <  7r/4  the  series  in  (b)  or  (c)  converges 
rapidly,  and  only  a  few  terms  need  to  be  computed. 

Having  found  the  value  of  the  sine  and  the  cosine,  the  log- 
arithmic sine  and  cosine  can  be  obtained  from  a  table  of 
logarithms  of  numbers.     The  logarithmic  tangent  can  then  be 


SIMPSON'S  METHOD  117 

found  by   subtracting   log  cos   from   log  sin,   and  log  cot  by 
subtracting  log  sin  from  log  cos,  or  log  tan  from  0. 

Example.     Compute  sin  11°  12'  23''  and  cos  11°  12'  23". 

By  §66, 

11°  =  (0.01  745  329  252  x  11)  radian  =  0.19  198  621  772  radian. 
12'  =  (0.00  029  088  821  x  12)       "      =  0.00  349  065  852       " 
23"  =  (0.00  000  484  814  x  23)      "      =  0.00  Oil  150  722       " 
.-.  0  =  11°  12' 23"  =0.19  558  838  346      " 

Substituting  this  value  for  <p  in  (b)  and  (c),  from  the  first  three  terms 
we  obtain 

0  =  0.19  558  838  1  =  1.00  000  000 

05/(5=  .00  000  238  0V|4=  .00  006  098 

.19  559  076  1.00  006  098 

08/13=    .00  124  703  0V|2=    .01912  741 

.-.  sin  0  =    .  19  434  373  .-.  cos  0  =    .98  093  357 

The  sine  is  correct  to  five  places  and  the  cosine  to  four  places. 

84.  Simpson's  methpd  of  computing  a  trigonometric  table  is  the 
following : 

Suppose  the  table  is  to  be  at  intervals  of  V. 

Putting  (71  -  1)  V  for  A  and  1'  for  B  in  (1)  and  (3)  of  §  40, 
we  obtain 

smnl'  =  2  cos  1'  sin (n  -  1)1' -  sin (n  -2)1'.        (1) 
cos  71 1'  =  2  cos  1'  cos  (n  -  1)  1'  -  cos  (n  -  2)  1'.        (2) 

Putting  30°  for  A  in  (1)  and  (4)  of  §  40,  we  obtain 

sin  (30°  -{-B)  =  GosB-  sin  (30°  -  B).  (3) 

cos  (30°  -\-B)  =  cos  (30°  -B)-  sin  B.  (4) 

Calculate  sin  1'  and  cos  1'  as  in  §  82  or  by  the  series  in  §  83. 

Then  giving  to  n  in  (1)  and  (2)  the  values  2,  S,  4,  etc.,  suc- 
cessively, we  obtain  the  sines  and  cosines  of  angles  up  to  30° 
at  intervals  of  1'. 

E.g. ,  when  n  =  2,  (1)  and  (2)  become 

sin  2'  =  2  cos  1'  sin  1'  —  sin  0'  =  2  cos  I'sin  1", 
and  cos  2'  =  2  cos  V  cos  1'  —  cos  1'  =  2  cos^  1'  —  1,  •  •  •, 


118  PLANE  TRIGONOMETRY 

To  obtain  the  sines  and  cosines  of  angles  from  30°  to  45°, 
we  give  to  B  in  (3)  and  (4)  the  values  V,  2',  3',  etc.,  succes- 
sively, making  use  of  the  results  previously  obtained  by  (1) 
and  (2). 

E.g.,  when  B  =  Y,  (3)  and  (4)  become 

sin  30°  r  =  cos  V  -  sin  29°  59', 
and  cos  30°  V  =  cos  29°  59'  -  sin  1',  ■  •  • . 

85.  The  hyperbolic  functions.  One  form  of  the  exponential 
series  is 

/j%Z  /ytO  /yi4  /y»5 

^'-^  +  -  +  l2  +  i  +  |  +  |  +  -'  W 

where  e  is  the  base  of  natural  logarithms. 

For  the  proof  of  (d),  see  §  95  in  Taylor's  Calculus,  or  §  326 
in  the  College  Algebra. 

Putting  a;  =  1,  we  obtain  e  =  2.718281. 

The  functions  (e*  —  e~^)/2  and  (e^  +  e"^)  /2  are  found  to  have 
properties  analogous  to  those  of  sin  x  and  cos  x,  and  to  have 
to  the  hyperbola  relations  analogous  to  those  which  sin  x  and 
cos  X  have  to  the  circle ;  for  these  reasons  the  first  has  been 
named  the  hyperbolic  sine  of  x  {sink  x),  and  the  second  the 
hyperbolic  cosine  of  x  (cosh  x).     Thus  we  have 

sinh  X  =  (e^  —  e-*)/2, 


coshx  =  (e^  4-e-^)/2.  '  "-     -" 

Following  the  analogy  of  the  trigonometric  functions,  we 
define  the  ratio  of  sinh  x  to  cosh  x  as  the  hyperbolic  tangent  of 
X  {tanh  x). 

Thus,  tanh  x  =  ^^  =  ^\~  ^'\'  [38] 

cosh  x      e^  +  e~^  *-     -' 

The  reciprocals  of  tanh  a?,  cosh  a-,  sinh  x  are  called  respec- 
tively the  hyperbolic  cotangent  of  x  {coth  x),  the  hyperbolic 
secant  ofji  (sech  x),  and  the  hyperbolic  cosecant  ofx  (csch  x). 


HYPERBOLIC  FUNCTIONS  119 

86.  The  relations  between  the  hyperbolic  functions  are  in  gen- 
eral analogous  to,  and  sometimes  the  same  as,  the  corresponding 
relations  between  the  trigonometric  functions. 

E.g.,         (e-  +  6-^)2/4  -  {^  -  e— )V4  =  1. 
.-.  cosh2  X  —  sinh^  x  =  1. 

Again,  cosh  (—  x)  = =  cosh  x, 

g — X  g.r 

and  sinh(— x)=  =— sinhx. 

Also,  smh  2  X  =  — — —  =  2 

2  2  2 

.-.  sinh  2  X  =  2  sinh  x  cosh  x. 

Again,  sinh  x  cosh  y  =  (e*  —  e-*)  (eJ'  -|-  e-'/)/4 

=  (e*  +  y  -  e-^+J'  +  e*-y- e-^-y)/4,    (1) 

and  cosh  x  sinh  y  =  (e*  +  e-*)  (e^  —  e-^)  /4 

^(ex+y  _|.  e-x+y  _  ea;-y  _  g-ar-y)/4_     (2) 

Adding  (1)  and  (2),  we  obtain 
sinhx  cosh  2/  -(-  cosh  x  sinh  y=  [e^  +  v  — e-<^  +  J')]/2 
=  sinh  (x  +  y). 

Example.  Prove  sinh  (x— y)  =  sinh  x  cosh  y  —  cosh  x  sinh  y. 
cosh  (x  4-  y)  =  cosh  x  cosh  y  +  sinh  x  sinh  y. 
cosh  (x  —  y)  ^  cosh  x  cosh  2/  —  sinh  x  sinli  y. 
The  notation  for  the  inverse  hyperbolic  functions  is  the  same  as  that 
for  the  inverse  trigonometric  functions. 

EXERCISE    XXVm 

Prove  each  of  the  following  identities  ; 

1.   tanh2  X  +  sech2  x  =  1 .  2.    cosh*  x  +  cosh*  x  =  1. 

„    ^     ,  ,     ,     ,       tanh  X -I- tanh  y 

3.  tanh  {x-\-  y)  = ^  • 

1  +  tanh  X  tanh  y 

4.  sinh  3  X  =  3  sinh  x  +  4  sinh'  x. 

5.  cosh  3  X  =  4  cosh^  x  —  3  cosh  x. 


6.  sinh- 1 X  =  cosh- 1  Vl  +  x*  =  tanh- 1  (x  /  VT+x*) . 

7.  tanh- ^ x  -I-  tanh- ' y  =  tanh- »  ^  "^^  • 

1  4-xy 


CHAPTEE   VIII 


COMPLEX  NUMBERS.     DE  MOIVRE'S   THEOREM 

87.    The   quality  units,  ±1,  ±  V^^Il.      Let  ABA'B'  be  a 

circle  whose  radius  OA  is  1  unit  in  length. 
Then  if  OA  =  the  unit  +1,  OA'  =  the  unit  "1. 
Now  OA  X  (~1)=0A';  that  is,  if  OA  is  multiplied  by  "1, 

OA  is  reversed  in  direction  and 
becomes  OA'.  Suppose  that  OA 
reverses  its  direction  by  revolv- 
ing about  O  in  the  plane  of  the 
figure;  then,  as  ~.l  =("'"V—  1)^ 
\j  or  ("V^)2,  it  follows  that  OA 
will  be  reversed  in  direction  if  it 
is  multiplied  twice  in  succession 
by  either  "^  V—  1  or  ~ V— 1. 
Hence  it  may  be  assumed  that 
the  effect  of  V—  1  or  ~  V—  1 
as  a  multiplier  will  be  to  revolve 

OA.   through  7r/2.      Suppose  that  "^V— 1,  as  a  multiplier, 

revolves  OA  in  the  positive  direction,  or  counter-clockwise ; 

then  "V—  1  will  revolve  OA  in  the  negative  direction,  or 

clockwise. 

That  is,    OA  x  "^  V^  =  OB,  OA  x  "  V^  =  OB'. 

Putting  +1  for  OA  and  assuming  the  commutative  law,  we 

have  J J 

OB  =  V^^,         OB'  =    V~  1. 

For  brevity  we  shall  denote  V—  1  by  i  and  ~  V— 1  by  ~i. 

The  quality  units  "•"!  and  ~1  are  often  called  rea^  units,  and 

arithmetic  multiples  of  these  units  are  called  real  numbers. 

120 


J 

5_ 

/"I 

\ 
\ 

\ 

\ 

/             + 

t'l   t  -1 

•"1    ^       1 

'!         0 

1 

\   ^r 

_^^^ 

Fig.  55 


QUALITY  UNITS  121 

The  quality  units  i  and  ~i  are  called  imaginary  units,  and 
arithmetic  multiples  of  these  units  are  called  imaginary 
numbers. 

Observe  that  +i  and  -i  or  +1  and  -1  include  both  the  idea  of  the 
arithmetic  1  and  that  of  oppositeness  to  each  other. 

Since  +lx+l  =  +l  or— Ix— 1  =  +1,  +1  or— 1  is  lYs  own  reciprocal. 

Since  +i  x  —i  =  +1,  +i  and  —i  are  reciprocals  of  each  other. 

Hence  i  and  —  i  are  both  opposites  and  reciprocals  of  each  other. 

Since  {—  i)^  =  i^  =  —1,  the  square  of  either  imaginary  unit  is— I. 

Also,  (+i)3  =  —  i  and  (-i)^  =  i;  that  is,  the  cube  of  either  imaginary 
unit  is  equal  to  the  other. 

Again,  ("i)*  =  i*  =  +1 ;  that  is,  the  fourth  power  of  either  imaginary 
unit  is  +1. 

88.  A  directed  line  or  a  directed  force  is  a  line  or  force  whose 
value  includes  not  only  its  magnitude  but  also  its  direction. 
E.g.,  OA,  OB,  OA',  OB',  A^,  BB',  in  fig.  55,  are  directed  lines. 

A  directed  line  is  often  called  a  vector. 

Two  directed  lines  or  forces  are  equal  when  they  have  the 
same  length  or  size  and  the  same  direction. 

Hence,  if  two  vectors  are  equal  and  their  origins  coincide, 
their  ends  also  will  coincide. 

If  a  directed  line  or  force  is  parallel  to  one  of  two  perpendicular  lines, 
as  ADA'  and  BOB'^  in  fig.  55,  we  can  express  both  its  magnitude  and  its 
direction  by  a  real  or  an  imaginary  number.  To  enlarge  our  number 
concept  so  that  we  can,  by  the  sum  of  a  real  and  an  imaginary  number, 
express  the  magnitude  and  direction  of  any  directed  line  or  force  which 
is  parallel  to  any  line  whatever  in  a  given  plane,  as  A  OB  in  fig.  55,  we 
need  the  principle  of  vector  addition  given  below. 

89.  In  fig.  55,  we  have 

OA'  -i-AM  =  0A,  OB  -\- BB'  =  OB', 

OA   +  AA'  =  OA',  'OB'  -\- B^  =  OB. 

Each  of  these  identities  illustrates  the  following  law  of 
vector  addition : 


122 


PLANE  TRIGONOMETRY 


Fig.  56 


If  the  end  of  one  vector  is  the  origin  of  a  second  vector,  the  sum 
of  these  two  vectors  in  its  simplest  form  is  the  vector  extending 
from  the  origin  of  the  first  vector  to  the  end  of  the  seco7id  vector. 

Thus,  in  its  simplest  form,  the  sum  of  the  vectors  AB  and 
BC  in  fig.  56  is  the  vector  AC. 

That  is,  AB-\-  BC  =  AC.  (1) 

The  meaning  of  (1)  is  that  transference  from  ^  to  -B  fol- 
lowed by  transference  from  j5  to  C  is 
identical  in  result  with  transference  from 
A  to  C. 

Or,   a  point  P  moving  from  A  to   C 
along  AC  goes  the  distance  AB  in  the 
direction  AB  plus  the  distance  BC  in  the 
direction  BC. 
Again,  if  two  directed  forces  are  represented  by  the  vectors 
AB  and  BC,  by  the  parallelogram  of  forces  we  know  that 
their  resultant  (i.e.  their  sum  in  its  simplest  form)  is  repre- 
sented by  the  vector  AC. 

Identity  (1)  is  absurd  when,  as  in  Geometry,  we  consider 
only  the  length  of  lines  ;  it  becomes  true  when,  and  only  when, 
the  value  of  each  line  includes  both  its  length  and  its  direction. 

90.  Complex  numbers.  If  a  and  b  are  real  numbers,  then 
a  -f-  ib  is  called  a  complex  number,  i.e.  a  complex  number  in 
the  common  form  is  the  sum 
of  a  real  and  an  imaginary 
number. 

•  To  construct  a  +  ib,  lay  off 
OM  =  a  and  MP  =  ib  ;  then 
a-\-ib  =  OM  +  MP  =  OP. 

Thus  the  vector  OP  repre- 
sents the  complex  number 
a  4-  ib,  or  a  -\-  ib  is  the  7iumerical  measure  of  the  vector  OP. 


COMPLEX  NUMBERS  123 

Take  OH  equal  to  1  unit  in  length  and  draw  HR  ±  OM. 
Let  r  =  the  arithmetical  value  of  V a^  +  ^^-  (1) 

Then,  from  the  right  triangle  OMP,  we  know  that 

r  =  the  number  of  units  in  OP. 

From  the  similarity  of  the  triangles  ORH  and  OMP,  we 
have 

'or  =  a/r,  RH  =  ib/r. 

Hence  OH  =  OR  +  RH  =  a/r-{-  ib/r. 

That  is,  a/r  -{•  ih  jr  denotes  OH,  the  unit  vector  of  OP. 
r  is  called  the  arithmetic  value,  or  modulus,  of  the  complex 
number  a  +  ib,  and  a /r  -{-  ib /r  \t^  quality  unit. 

If     Z  il/OP  =  <^,  then  a/r  =  cos  <^,  ^>/r  =  sin  <^.  (2) 

.'.  a/r  -{-  ib /r  =  Qos  <fi  -[-  i  sin  <^.  (3) 

.*.  a  -\-  ib  =  (a/r  +  ib/r)  r  =  (cos  <j>  -{-  i  sin  <^)  r.  (4) 

(cos  <f>  +  i  sin  <f>)  r  is  the  trigonometric  form  of  a  complex 
number  in  which  cos  4>  -\-  i  sin  <^  is  the  quality  unit  and  r  the 
modulus. 

Either  {a/r  -\-  ib /r)r  or  (cos  4>  ■\-  i  sin  <f>)r  in  (4)  is  called 
the  type  form  of  the  complex  number  a  +  ib. 

Hence  a  complex  number  in  the  type  form  is  an  arithmetic 
multiple  of  a  quality  unit. 

The  angle  <^  is  called  the  angle,  or  amplitude,  of  the  complex 
number  a  +  ib.  Between  —  it  and  +  tt  there  is  one,  and  only 
one,  value  of  <^  which  will  satisfy  equations  (2).  This  value 
of  <^  is  called  its  principal  value. 

If  <^'  denotes  the  principal  value  of  <j>,  the  general  value  of 
<^  is  2  niT  +  <^',  where  n  is  any  integer.  Thus  the  angle,  or 
amplitude,  of  a  complex  number  is  many-valued. 

When       6  =  0,  a  H-  t^  =  a,  a  real  number  ; 
when       a  =  0,  a  +  t6  =  ib,  an  imaginary  number. 


124  PLANE  TRIGONOMETRY 

Thus  a  -f  ih  includes  reals  and  imaginaries  as  particular 
cases. 

Ex.  1.  Reduce  the  complex  number  —  1  —  V  — 3  to  the  trigonometric 
type  form. 

Here  a  =  —  1,     &  =  —  V3- 

.-.  r=Vn-3  =  2.  by  (1) 

Hence  cos  0  =  -1/2,     sin  </>  =  -  V3/2.  by  (2) 

Since  cos0  and  sin  0  are  both  — ,  0  is  in  the  third  quadrant ;  hence 
its  principal  value  is  —  2  ;r/3. 

...  _i  _Vir3  3::[cos(-2  7r/3)  +  isin(-2  7r/3)]  -2. 

Hence  the  arithmetic  value,  or  modulus,  of  —  1  —  V—  3  is  2,  and  its 
quality  unit  is  represented  by  a  unit  vector  which  makes  the  angle 
—  2;r/3  with  the  vector  representing  +1. 

Ex.  2.  Reduce  the  complex  number  2  —  V—  5  to  the  algebraic  type 
form. 

Here  a  =  +  2,    6  =  -  V5 ;    •••  r  =  3. 

Ex.  3.     Reduce  the  unit  sin  0  —  i  cos  0  to  the  trigonometric  type  form. 

sin  0  —  i  cos  0  =  cos  (0  -  90°)  +  i  sin  (0  —  90°). 
Thus  the  angle  of  the  quality  unit  sin  0  —  i  cos  0  is  0  —  90°. 


EXERCISE   XXIX 

Represent  graphically  and  reduce  to  the  type  form  : 

1.  i+VZTi.  5.    _^3  +  i  9.    _6-i8. 

2.  1  -  V^.  6.    -  V3  -  i.  10.    -  V5  -  *  Vll- 

3.  —  1  —  i.  7.   3  — i4.  11.   sin0  +  icos0. 

4.  _  1  +  v^ITs.         8.    -  3  +  i  2.  12.    -  sin  ^  +  i  cos  d. 
13.    By  constructing  the  sum  in  each  member,  show  that 

(a  +  ih)  +  («  +  iy)  =  (a  +  iy)  +  (x  +  i6)  =  (a  +  x)  +  i  (6  +  y), 

and  thus  prove  geometrically  that  the  commutative  and  associative  laws 
of  addition  hold  true  for  complex  numbers. 


GENERAL  QUALITY  UNIT 


125 


91.    General  quality  unit  cos  <|)  +  i  sin  <|>. 
Let  ABA'  be  a  circle  whose  radius  is  1,  and  let  OA  =  the 
unit  +1. 

B 


Then  if  <^  =  ^  OP,  cos  <^  +  t  sin  <^  =  OP ; 

if  <!>  =  A  OPi',  cos  </>  +  ^  sin  </»  =  OP^' ; 

ii  (f>  =  A  OPi,  cos  <f>  +  i  sin  <^  =  OPi ; 

if  <^  =  yI  op',  cos  <f>-^ism<f>  =  OP'. 

That  is,  the  quality  unit  cos  <f}  +  i  sin  <f}  is  represented  by 
the  U7iit  vector  which  makes  the  angle  <^  with  the  vector  repre- 
senting "♦"!. 

The  following  important  particular  cases  of  cos  <^  +  i  sin  <^ 
should  be  carefully  noted : 

When  <^  =  0,  cos  <f>  -{- i  sin  <f>  =  cos  0  +  *  sin  0  =  + 1. 

When  <f>  =  ±  7r/2, 

cos  <^  +  t  sin  <^  =  cos(±  7r/2)  +  i  sin(±  7r/2)  =  *i. 
When  <f>  =  ±7r, 

cos  <^  +  t  sin  <^  =  cos  (±  tt)  +  i  sin  (±  tt)  =  ~1. 

These  results  are  evident  also  from  the  figure. 


126  PLANE  TKIGONOMETRY 

92.  Product  of  quality  units.  By  the  distributive  law  of 
multiplication,  we  obtain 

(cos  <^i  +  i  sin  <^i)  (cos  cf>2  +  ^  sin  c^g) 
=  cos  <^i  cos  <fi2  +  i  cos  <^i  sin  <^2  +  *'  sin  <^i  cos  <^2  +  **^  sin  <^i  sin  <f>2 
=  cos  <f)i  COS  <^2  —  sin  <^i  sin  <f>2  +  i  (sin  </>i  cos  <^2  +  cos  <^i  sin  <f>2) 
=  cos  (<^i  -\-  cji2)  +  i  sin  (cf>i  +  cfj^).  §  34 

Hence,  in  general,  we  have 

(cos  <j)i  +  i  sin  ^j)  (cos  <|)2  +  i  sin  ^2)  •  •  •  (cos  <|>n  +  i  sin  <|)n) 
=  C08((|)i  +  4)2  +  •  ■  •  +  <t>n)  +  i  sin(<t)i  +  <t)2  +  •  •  •  +  <t)„).      [39] 

That  is,  the  product  of  two  or  more  quality  units  is  a  quality 
unit  whose  angle  is  the  sum  of  the  angles  of  the  factors. 

Observe  that,  to  multiply  cos  ^i  +  i  sin  0i  by  cos  (p2  +  i  sin  <p2,  we  add 
the  angle  02  of  the  multiplier  to  the  angle  <pi  of  the  multiplicand. 

Again,  to  obtain  the  multiplier  cos  02  +  *  sin  02  from  the  primary  unit 
cos  0  +  i  sin  0,  we  add  the  angle  02  of  the  multiplier  to  the  angle  0  of  the 
primary  unit. 

Hence  the  product  is  obtained  by  doing  to  the  multiplicand  just  what 
is  done  to  the  primary  unit  to  obtain  the  multiplier. 

93.  De  Moivre's  theorem. 

(i)      If  ^  =  cf>,  =  ct>,  =   ...  =  <t>n, 

then,  from  [39]  of  §  92,  we  obtain 

(cos  </>  +  i  sin  <^)"  =  cos  ncfi  4-  i  sin  w<^.  (1) 

That  is,  the  nth  power  of  a  quality  unit  is  a  quality  unit 
whose  angle  is  n  times  the  angle  of  the  base. 

(ii)    Taking  the  wth  root  of  each  member  of  (1),  we  obtain 

(cos  n<fi  +  i  sin  7i<f>y^'^  =  cos  <f>  +  i  sin  <^. 
Putting  <f>  for  nct>  and  therefore  <f>/n  for  <^,  we  have 

(cos  <^  +  ^  sin  <f>y^"  =  cos  (<t>/n)  +  i  sin  (<f>/n).       (2) 


DE  MOIVRE'S  THEOREM  127 

Let  5  be  a  positive  integer,  then,  from  (2),  we  obtain 
(cos  <^  +  *  sin  c^)"/"  =  [cos  {4> /n)  +  i  sin  (<^/w)  J 

=  cos  (s<t>/n)  +  i  sin  (s<f>/7i).      (3) 
(iii)    By  §92, 

(cos  <f>-{-i  sin  <^)  [cos  (—  <f>)  +  i  sin  (—  <^)]  =  cos  0  +  i  sin  0  =  +1. 
Hence 

cos  (—  <^)  +  1  sin  (—  <^)  is  the  reciprocal  of  cos  <f>  -{-  i  sin  <^. 
That  is,  (cos  (f>  -{- i  sin  <^)-^  =  cos  (—  <^)  +  i  sin  (—  <^).       (4) 
Let  p  be  any  positive  integer  or  fraction.     Then,  from  (4), 
(cos  <t>  -{-  i  sin  <^)~^  =  [cos  (—  <^)  +  i  sin  (—  <f>)Y 

=  cos  (—  p<f>)  +  i  sin  (—  jxf}).  (5) 

Erom  (1),  (3),  and  (5)  it  follows  that  for  any  commensura- 
ble real  value  of  n,  we  have 

(cos  <|)  +  i  sin  (j))"^  =  cos  n<|)  +  i  sin  n<|).  [40] 

Formula  [40]  is  called  De  Moivre's  theorem. 

CoR.  1.     From  (2),  cos  (<^/n)  +  i  sin  (<^/n)  is  one  of  the  nth 
roots  of  cos  <^  +  i  sin  <^. 
CoR.  2.     By  §  28, 

cos  (—  <^)  +  i  sin  (—  <j>)  =  cos  <^  —  i  sin  <^. 

Hence,  by  (4),  cos  <^—i  sin  <^  is  the  reciprocal  of  cos  </>  +  ^  sin<^. 

That  is,  the  conjugate  quality  ^tnits  cos  <^  +  i  sin  <^  and 
cos  <f>  —  i  sin  <^  are  reciprocals  of  each  other. 

In  fig.  58  observe  that  OP  and  OPi  represent  reciprocal,  or 
conjugate,  quality  units,  while  OP  and  OP'  represent  opposite 
quality  units.  When  OP  coincides  with  OB,  OP^  and  OP'  both 
coincide  with  OB';  this  illustrates  that  the  reciprocal,  or  con- 
jugate, of  i  is  also  the  opposite  of  i.  When  OP  coincides  with 
OA  or  OA',  OPi  does  also;  that  is,  either  +1  or  ~1  is  its  own 
reciprocal. 


128  PLANE  TRIGONOMETRY 

94.    To  divide  by  the  quality  unit  cos  <f>  +  i  sin  <^,  we  multiply 
by  its  reciprocal  cos  (—  <^)  +  i  sin  (—  <f>),  or  cos  <f)  —  i  sin  </>. 

Ex.  1.      — ^— ^  =  (cos  01  +  i  sin  0i)  [cos  (-  02)  +i  sin  (-  02)] 

cos  02  +  *  sin  02 

=  cos  (01  —  02)  +  i  sin  (0i  —  02). 

Ex.  2.     (^os'^  +  ^s^^'^)'  =  (cos  0  +  i  sin  0)5  (cos  ^  +  i  sin  ey 
(cos^-isin^)7      ^       "^  ^  "^^  ^         ^  ' 

=  (cos  5  0  +  i  sin  5  0)  (cos  7  ^  +  i  sin  7  ^) 

=  cos  (5  0  +  7  ^)  +  i  sin  (5  0  4-  7  6). 


EXERCISE  XXX 

Prove  each  of  the  following  identities  : 

1.  (cos0  +  isin0)3  = +1,   [cos(2  ;r/3)+ isin(2;r/3)]3  = +1, 
[cos(4;r/3)+  isin(4;r/3)]3  =:  +1. 

Hence  each  of  these  three  quality  units  is  a  cube  root  of  +1. 

2.  [cos(7r/5)±isin(7r/5)]6  =  -l,  [cos(3  7r/5)±  isin(3  7r/5)]5  =-1, 
(cos  TT  lb  i  sin  7t)^  =  -1. 

Observe  that        cos  tt  +  i  sin  tt  =  cos  rt  —  ismTt. 

3.  [cos  {7t/6)±i  sin  {7t/6)f  =  [cos  {7t/2)±i  sin  {7t/2)Y 

=  [cos(5;r/6)±isin(5  7r/6)]6=-l. 

4.  What  is  proved  by  examples  2  and  3  ?     Illustrate  the  meaning  of 
examples  1,  2,  3  by  vectors. 


Express  as  a  quality  unit  each  of  the  following : 

_     (cos  ^  +  i  sin  0)^  _     (cos  ^  -  i  sin  6)^^^ 

5.    •  o. • 

(cos  0  —  i  sin  0)^  (cos  0  —  i  sin  0)^  /s 


(cos  ^  +  i  sin  6)  (cos  0  +  i  sin  0) 
(cos  j3  -}-  i  sin  /3)  (cos  7  +  *  sin  7) 

[cos  (7r/6)  -  i  sin  (Tr/G)]"/^  (cos  0  +  t  sin  0)^ 

[cos(;r/6)  +  isin(;r/6)]i/2*  *    (sin  ^  +  i  cos  ^)6  ' 


COMPLEX  NUMBERS  129 

10.   By  De  MouTe's  theorem  prove  identities  (1)  and  (2). 

cos  n0  =  cos«  <p ^^ COS" -2  <p  sin^  <p 

n(n-l)(n-2)(n-3)        „    4^.4^  «, 

-\ — ^ ^-^ '-^ ^cos"-^0sin*0  —  •  •  •.  (1) 

,      .    ■        n{n-l){n-2)      „    .      .  - 
Sin  n<p  =  n  cos"-i  0  sin  0 ^^ -^ cos«-3 0  sin^ 0 

7i(n-l)(n-2)(n-3)(n-4)       „    .^   .  .^  ,ov 

-I — 5^ LI L!^ Li L cos"-^ 0 sm^ 4>  —  '•  '.    (2) 

cos  710  +  i  sin  n0  =  (cos  <p  +  i  sin  0)« 

=  cos"  0  +  t  -  cos»»-i  0  sin  0  +  i^  — ^ cos»-2  0  sin^  0 

1  [2 

+  ^3^(^-^)O^-^)cosn-3  0sina0  +  ....  (3) 

Substituting  for  i^,  i^,  ...  their  values,  and  equating  the  real  parts 
and  the  imaginary  parts  in  (3),  we  obtain  (1)  and  (2). 

95.  Products  and  quotients  of  complex  numbers.  Multiplying 
by  a  quality  unit  affects  only  the  qimlity  of  the  product,  and 
multiplying  by  an  arithmetic  number,  or  modulus,  affects  only 
the  size  of  the  product;  whence  the  result  is  evidently  inde- 
pendent of  the  order  of  these  operations.  Therefore  the  com- 
mutative and  associative  laws  of  multiplication  hold  for  the 
moduli  and  quality  units  of  complex  numbers. 

Hence  we  have  the  following  theorems : 

(i)  The  product  of  two  or  more  complex  numbers  is  equal  to 
the  product  of  their  quality  units  into  the  product  of  their  moduli. 

(ii)  The  quotient  of  two  complex  numbers  is  equal  to  the  quo- 
tient of  their  quality  units  into  the  quotient  of  their  moduli. 

(iii)  The  mth  power  of  any  complex  number  is  equal  to  the 
rath  power  of  its  quality  unit  into  the  mth  power  of  its  modulus, 
where  m  is  any  real  number. 


130  PLANE  TRIGONOMETRY 

96.    The  quality  unit  cos  ^  +  ^  sin  <^  has  q,  and  only  q,  unequal 
j\th  roots. 

If  n  is  any  integer,  then,  by  §  21,  we  have 

cos  <f>  -\-  i  sin  <^  =  cos  (2  titt  +  <^)  +  i  sin  (2  ^^tt  +  <^). 
.-.  (cos  4>  +  i  sin  <^)i/«  =  [cos  (2  ?i7r  +  (^)  +  i  sin  (2  titt  +  <^)]^  " 

=  cos ^^-^H-isin ^^-^-      (1) 

If  in  (1)  we  give  to  n  the  values  0, 1,  2,8,  A,-  ■  ,  q  —  1,  in  suc- 
cession, we  obtain  the  following  q  qth  roots  of  cos  <^  +  i  sin  <f> : 

9z  =  0,  cos  (<i>/q)  -\-  i  sin  {(^/q), 

.                           27r-h<t>  ,    .    .    2  7r  +  (^ 
71  =  1,  COS +  t  Sin J 

Att  -{-  d>       ..47r-f<f) 
?z  =  2,  COS +  I  Sin ? 


(2) 


The  angles  of  any  two  of  the  roots  in  (2)  differ  by  less  than 
2  TT ;  hence  no  two  of  these  angles  can  have  equal  cosines  and 
equal  sines.     Therefore  the  q  qth  roots  in  (2)  are  unequal. 

If  we  give  to  n  the  values  q,  q  -\- 1,  q  -\-  2,  -  •  -,  ot  —  1,  —  2, 
—  3,  •  •  •,  we  obtain  g-th  roots  equal  to  those  in  (2). 

Hence  cos  <f>  -{-  i  sin  ^  has  q,  and  only  q,  unequal  qth  roots. 

97.    Any  complex  7iumber  has  q,  and  only  q,  unequal  qth  roots. 
[(cos  cf>  -\-  isin  <^)  rj^'^  =  (cos  <^  +  i  sin  <^)  ^  /-^r^/?.         (1) 

The  quality  unit  cos  <t>  -\-  i  sin  <^  has  q,  and  only  q,  unequal 
qth  roots,  and  the  arithmetic  number  r  has  only  one  qth  root ; 
whence  the  second  member  of  (1)  has  q,  and  only  q,  unequal 
values. 

Hence  to  find  the  q  qth  roots  of  any  complex  number,  reduce 
the  number  to  the  type  form,  find  the  q  qth  roots  of  its  quality 
unit,  and  multiply  each  by  the  ^-th  root  of  the  modulus. 


ROOTS  OF  COMPLEX  NUMBERS  131 

Observe  that  an  algebraic  number  has  q  unequal  qth  roots  because  its 
quality  unit  has  q  unequal  qth  roots. 

Ex.  1.     Find  the  three  cube  roots  of  —  27. 

-27  =  (-!)•  27  =  (cosTT  +  isin^r)  -27. 

.-.  V-  27  =  (cos  TT  +  i  sin  TT)  1  /-^  •  3. 

,  ,    .   .       ,,,o  2n7t  +  Tt  ^    .   .    2n7r -\- 7t 

(cos  7t  +  ism  7t)^'^  =  cos h  I  sin —  ; 

^  '  3  3 

when  n  =  0,  =cos(;r/3)  +  isin(;r/3)  =  1/2  +  i  V3/2; 

when  71  =  1,  =  cos  ;r  +  i  sin  tt  =  —  1  ; 

when  n  =  2,  =cos(5  7r/3)  +  isin(5  7r/3)  =  1/2  -i^S/2. 


Hence  V_  27  =  -3,  (1  /2  ±  i  V3/2)  •  3. 

Ex.  2.     Find  the  four  fourth  roots  of  8  +  8  V^. 

8  +  8  V3T3  =  (1  /2  +  i  V3/2) .  16  =  [cos  {tt/S)  +  i  sin  (7r/3)]  •  16. 
...  V8  +  8  V33  =  [cos(;r/3)  +  isin(7r/3)]  i/* -.2. 

r       /     /o^   ,    •   •    /     /oMi/d  2n7r  +  7r/3  ,    .   .    2n7t  +  7t/3 

[cos(7r/3)  +  tsm{7t/d)Y^*  =  cos —  +  ism —  ; 

4  4 

when  n  =  0,  =  cos  (7r/12)  +  isin  (;r/12); 

when  n  =  1,  =  cos  (7  ;r/ 12)  -f  isin  (7  ;r/12); 

when  n  =  2,  =  cos  (13  ;r  / 12)  +  i  sin  (13  7t  / 12) 

when  n  =  3,.  =  cos  (19  ;r/12)  +  isin  (19  ;r/ 12).  ) 


(1) 


Multiplying  each  of  the  four  units  in  (1)  by  2,  we  obtain  the  four 
required  fourth  roots. 

EXERCISE  XXXI 
Find  all  the  values  of : 

1.  11/3.                             5.    161/4.  9.  (4V3  +  i4)'/3. 

2.  (-1)1/3.                       6.    321/5.  10.  (i_  4^3)1/4. 

3.  11/6.                              7.    (_243)i/\  11.  (l+\/Zl)i/6, 

4.  (- l)i/«.                        8.    {-iy/^  12.  (V3-V^)2/5. 

13.    Solve  the  equation    x*  -  x^  +  x2  -  x  +  1  =  0.  (1) 

Multiply  by  (X  +  1),  x^  +  1  =  0,  or  x  =  (-  l)i/5. 


132  PLANE  TEIGONOMETRY 

Multiplying  by  jc  +  1  introduces  the  root  -  1,  the  other  four  of  the 
five  fifth  roots  of  —  1  are  the  roots  of  equation  (1). 

14.  Solve  the  equation  x^^  —  1  =  0. 
Factor,             (x^  -  1)  {x^  +  1)  {x»  -  i)  {x^  +  i)  =  0. 

The  twelve  roots,  V/%  (-  l)i/3,  ii/3^  {-ly/^  are  readily  found  by 
De  Moivre's  theorem. 

15.  Solve  the  equation  x^  +  x*  +  x^  +  1  =  0. 
Factor,  (x*  +  1)  (x^  +  1)  =  0. 

16.  Solve  the  equation  x^  +  1  =  0. 

98.    Exponential  form  for  cos  cj)  +  i  sin  <|>.     If  in  series  (d)  of 
§  85  we  replace  x  by  i(l>,  we  obtain  the  series  (1). 

^  +  "l'  +  -\2'+-\f      IT      ^  (1) 

=  COS  <^  +  i  sin  </>.  by  (b),  (c),  §  83 

That  is,  series  (1)  is  equal  to  a  quality  unit  whose  angle  is  <f>. 
If,  as  is  suggested  by  (d),  we  define  c'*  as  an  exponential 
symbol  for  the  series  (1),  we  have 

e'^  =  cos  <^  +  i  sin  <j>.  (2) 

That  is,  €*''*  is  an  exponential  symbol  for  a  quality  unit  whose 
angle  is  <j>. 

Substituting  —  <l>  toi  <j>  in  (2),  we  obtain 

c"'*^  =  cos  <^  —  i  sin  <f>.  (3) 

From  (2)  and  (3),  e'*  and  e"~'*  denote  reciprocal  quality  units. 
From  (2)  and  §§  92  and  94,  it  follows  that 

^i<t)  _^  gi0  =  ^i<t>  ,^-i9  =.  ^ii<t>-0)^       (c**)^  =  C*^*. 


THE  EXPRESSION  e^  133 

That  is,  €**  obeys  the  fundamental  laws  of  exponents. 
The  unit  of  <j>  is  the  radian.     Why? 
Putting  1  for  <^  in  (2),  .we  obtain 

c*  =  cos  1  -h  ^'  sin  1. 

That  is,  c*  denotes  a  quality  unit  whose  angle  is  a  radian. 

Example.     What  does  e-*  denote  ?    6^  ? 

Observe  that  for  all  values  of  0  the  arithmetic  value  of  e'*  is  1. 

Hence,  as  0  varies  from  —  oo  to  +  oo,  e'*  varies  in  quality  only. 

Therefore,  e  in  e»<^  cannot  have  the  meaning  it  has  in  e^. 

Observe  that  the  numerical  measure  of  any  directed  line  or  force  in  a 
given  plane  can  be  expressed  in  the  form  e'*  •  a,  where  a  is  an  arithmetic 
number  which  gives  its  length  or  size  and  <p  is  some  angle  between  —  it 
and  It  which  gives  its  direction. 

Adding  and  subtracting  (3)  and  (2),  we  obtain 

COS  <^  = .  sin  4>  =  — — (4) 

The  values  in  (4)  are  known  as  Euler's  exponential  values 
of  cos  <^  and  sin  <^. 

Compare  (4)  with  [37]  in  §  85. 


CHAPTEK  IX 

MISCELLANEOUS  EXAMPLES 

EXERCISE  XXXII 

Express  all  the  angles  which  are  coterrainal  with  : 

1.    45°.  2.    132°  3.    -  35°.  4.    -  100°.  5.    it/^. 

Find  all  the  other  trigonometric  ratios  of  A  when  : 

6.  sin  ^  =  4/7.  8.    cos^  =  -3/8.         10.    secJ.  =  7/4. 

7.  tanJ.  =  3/2.  9.    cot^  =  -7/5.         11.    csc^  =  -5/4. 

12.  In  what  quadrant  is  A  in  each  of  the  examples  6-11  inclusive  ? 
Construct  A  in  each. 

In  terms  of  a  function  of  an  angle  less  than  45°  express  : 

13.  sin  94°.  16.    cot  320°.  19.    cos  (-175°). 

14.  cos  128°.  17.    sec  190°.  20.    tan  (-200°). 

15.  tan  215°.  18.    sin  (-75°).  21.    cot  (-300°). 
In  terms  of  each  of  the  other  functions  of  A  find  the  value  of : 

22.  sin^.  24.    tan^.  26.    sec  ^. 

23.  cos  J..  25.    cot  J..  27.    csc^. 

Identities 

Prove  each  of  the  following  identities  : 

28.  (tan  A  +  cot  ^ )  sin  ^  cos  ^  =  1. 

29.  (sec  A  —  tan  A)  (sec  A  +  tan  A)  =  1. 

30.  (esc  A  —  cot  A)  (esc  A  +  cot  A)  =  1. 

31.  (sin B  -cosB)^  =  l-2  sin  B cos B. 

32.  sin  B  +  cosB=y/2  cos  {B  -  7C/4). 

33.  sin  B  -  cos  B  =  -  V2  cos  {B  +  tt  /4). 

134 


^ 


IDENTITIES  135 

34.  sin  {A  -\-  7t/S)-{-  sin  (A  -  7t/3)  =  sin  A. 

35.  cos  {A  +  7r/6)  +  cos  {A  -  7t/6)=  V3  cos  A. 

36.  (cot  J.  +  tan  5)  /(tan  vl  +  cot  B)  =  cot  ^  tan  B. 

37.  1  -  tan*  A  =2  sec2  ^  -  sec*  A. 

38.  sec  ^/  (1  +  cos  B)  =  (tan  5  -  sin  B)/sm^  B. 

39.  sec2  A  csc2  4  =  tan2  A  +  cot2  J.  +  2. 

40.  tanB +  sec5  =  tan(jB/2  +  ;r/4). 

41.  (1  +  tan  B)/{1-  tan  B)  =  (cot  B -\- 1)  /  (cot  5  -  1). 

42.  sin  /I  /  (1  -f  cos  ^)  +  (1  +  cos  A) /sin  A  =2  esc  ^. 

43.  sec -3  A.  —  sin^  A  =  (cos  ^  —  sin  A)  (1  +  sin  A  cos  tI). 

44.  (sin  A  cos  E  +  cos  A  sin  5)2  -i-  (cos  A  cos  5  —  sin  J.  sin  5)2  =  1. 

45.  cot^  —  tan  ^  =  2cot  2-4. 

46.  sec  2  ^  =  sec2  A  /  {2  -  sec2  ^) . 

47.  2sec2^  =sec(^  +  ;r/4)sec(^  -  7r/4). 

48.  sin2^  =  2tan^/(l  +  tan2^). 

49.  2  sin  J.  +  sin  2  ^  =  2  sin^  A /{I-  cos  A). 

50.  Find  sin  ( J.  +  J5  +  C7)  in  terms  of  the  sine  and  cosine  of  A^  B,  C. 
Applying  formula  [7]  twice  and  [8]  once,  we  obtain 

sin(^  +  5  +  C)  =  sin(^  +  5)cos  C  +  cos(^  +  B)sin  G 
=  (sin  A  cos  B  +  cos  A  sin  B)  cos  C 

+  (cos  A  cos  5  —  sin  ^  sin  B)  sin  C 
=  sin  A  cos  J5  cos  C  +  cos  ^  sin  B  cos  C 

4-  cos  A  cos  5  sin  C  —  sin  J.  sin  B  sin  C       (1) 

It  A,  B,  C  are  the  angles  of  a  triangle,  sin  (A  +  B  +  C)  =  sin  180°  =  0. 
Hence,  from  (1),  we  obtain 
sin -4  cos  5  cos  C  +  cos^sin  J5cosC  +  cos^  cos  B  sin  C  =  sin  ^  sin  5  sin  C 

5i.    Applying  formula  [8]  twice  and  [7]  once,  prove 
cos  {A  -\-  B  -{-  C)  =  cos  A  cos  B  cos  C  —  cos  A  sin  B  sin  C 

—  sin  A  cos  2?  sin  C  —  sin  ^  sin  B  cos  C.  (2) 


136  PLANE  TRIGONOMETRY 

52.   Find  tan  {A  +  B  +  0)  in  terms  of  tan  A,  tan  5,  tan  C. 
Applying  formula  [11]  three  times,  we  obtain 

tan(^  +  ^  +  C)^^^^(-^  +  ^^  +  ^^"^ 
^  '      1  -  tan  (^  +  ^)  tan  C 

tan  A  +  tan  B 


1  —  tan  A  tan  B 


+  tanC 


tan  A  +  tan  B  ^      ^ 
tan  C 


1  —  tan  A  tan  B 
_   tan  J.  +  tan  B  +  tan  C  —  tan  A  tan  ^  tan  C        „, 
~  1  —  tan  A  tan  J5  —  tan  A  tan  C  —  tan  B  tan  C 
If  ^,  B,  O  are  the  angles  of  a  triangle,  tan  {A-\-B^-C)  =  tan  180°  =  0. 
Hence,  from  (3), 

tan  A  +  tan  B  +  tan  C  =  tan  A  tan  B  tan  C 

53.  Puttmg  ^  =  5  =  0  in  (1),  (2),  (3)  of  examples  50,  51,  52,  prove 

sin  3  ^  =  3  sin  ^  (1  -  sin2  J.)  -  sin2  J. 

=  3sin  J. -4sin3^,  (4) 

cos  SA  =  4  cos^  J.  —  3  cos  J.,  (5) 

^      o   .  _  3  tan  Jl  -  tan  3  ^ 

tan  3  ^  = — (6) 

1  -  3  tan2^  ^  ' 

Identity  (5)  is  useful  in  solving  cubic  equations.     See  example  112. 

54.  Writing  SA  =  2A  +  A,  prove  (4),  (5),  (6)  in  example  53  by  using 
[7],  [8],  [11],  [13],  [14],  and  [15]. 

55.  Substituting  ^  for  3  ^  in  (4)  and  (5)  of  example  53,  we  obtain 

sin  e  =  3  sin  (^/3)  -  4  sin3  (^/3), 
cos  ^  =  4  cos3  (d/Z)  -  3  cos  (d/S). 

Prove  each  of  the  following  identities  : 

56.  sin  4  J.  =  4  sin  ^  cos  ^  -  8  sin^  ^  cos  J. 

=  8  cos^  ^  sin  ^  —  4  cos  A  sin  A. 

57.  cos4^  =  l-8cos2^  +  8cos4^ 

=  l-8sin2J.  +  8sin4^. 

58.  cos  780°  =  1/2.  60.    cos  2550°  =  y/S/2. 

59.  sin  1485°  =  V2/2.  61.   sin(- 3000°)  =  -  V3/2. 

62,    tan  (-  2190°)  =  -  V3/3. 


EQUATIONS  137 

Equations 

In  what  quadrant  is  A  in  each  of  the  following  equations  ? 

63.  sin  J.  cos^  =  -  2/3.  65.    sec  ^  tan  J.  =  -  3. 

64.  sin  ^  tan  ^  =  4.  66.    cot  J.  +  3  sin  ^  =  0. 

67.  Solve  the  equation  sin  2  ^  =  2  cos  6.  (1) 

Substituting  for  sin  2  d  its  identical  expression  2  sin  6  cos  6,  from  (1)  by 
Algebra  we  obtain  the  equivalent  equation 

2  sin  ^  cos  ^  =  2  cos  6,  or  cos  0  (sin  ^  —  1)  =  0. 
From  cos  ^  =  0,  e  =  n7t  ±7t/2.  §70 

From  sin  ^  =  1,  d  =  UTt  +{- l)'^7t/2.  §69 

Hence  mt  ±  it  /I  includes  all  the  values  of  Q  in  (1). 

Solve  each  of  the  following  equations  : 

68.  cos2^  =  2sin^.  82.  2  sin-2  x  -  2  =  -  V2  cos  x. 

69.  cos  ^  =  sin  2^.  83.  cos2y4-2sin2?/ —  fsiny  =  0. 

70.  sin  ^  =  cos  2  ^.  84.  sin  ^  +  sin  2  ^  =  1  -  cos  2  6. 

71.  tan  A  tan  2  A  =2.  85.  cos  y  —  cos  2y  =  1. 

72.  cos  ^  +  cos  2  ^  =  0.  86.  sin  (45°  +  z)  +  cos  (45°  -z)  =  l. 

73.  cot  ^  tan  2  ^  =  3.  87.  sec  2  z  +  1  =  2  cos z. 

74.  4  cos  2  ^  +  3  cos  J.  =  1.  88.  cos  2  z  =  a  (1  -  cos  z). 

75.  sin  0  sec  2  ^  =  1 .  89.  tan  2  y  tan  y  =  1. 

76.  cot  ^  tan  2  ^  -  sec  2  d.  90.  sec  ^  =  2  tan  ^  +  1  /4. 

77.  sin  2  ^  =  3  sin2  ^  -  cos2  ^.  91.  sin-^x  +  sin-i  (x/2)  =  120° 

78.  sin  ^  +  cos  2  ^  =  4  sin2  0.  92.  sin- 1  z  +  2  cos- 1  z  =  210°. 

79.  sin  2  ^33  cos 4^.  93  tan-i  ?/ +  2  cot-iy  =  135°. 

80.  sec  X  +  tan  x  =  ±  y/S.  9  „ 

94.    tan- 1  -^ —  =  60°. 

81.  tan  X  +  2  V3  cos  x  =  0.  1  -  z2 

95.  tan-i  z  +  tan-i  2  z  =  tan-i  3  V3. 

96.  tan  x  +  tan  2  x  =  0. 


138  PLANE  TRIGONOMETRY 

97.  tan2  x  +  cot2  x  =  10/3.         99.   sin  A  +  cosA=  sec  A. 

98.  4  cos  2  ^  +  6  sin  6  =  5.         100.   sin  {6  +  30°)  sin  (d  -  30°)  =  1/2. 

Systems  of  Equations 

101.    Solve  for  r  and  6  the  system 

rsind  =  a,        (1) 


rcos^  =  6.         (2)  ^  ^^' 

Divide  (1)  by  (2),  tan  d  =  a/b,  or  6  =  tan-i(a/6). 

Square  (1)  and  (2)  and  add, 

r2(sin2^  +  cos2^)  =  a2  +  62. 
.-.  r  =  Va2  4.  52. 

102.  Solve  for  r,  0,  and  0  the  system 

r  cos  0  sin  ^  =  a,        (1)  ^ 

r  cos  0  cos  ^  =  6,         (2)  i-  (a) 

r  sin  0  =  c.         (3)  j 

Divide  (1)  by  (2),  tan  d  =  a/b,  or  ^  =  tan-i  (a/6).  (4) 

Square  (1)  and  (2)  and  add, 

r2  cos2  0  =  a2  +  62.  (5) 

From  (5),  r  cos  0  =:  VoM^.  (6) 

Divide  (3)  by  (6),  tan  0  =  c/Va2  +  62^  or  0  =  tan-i  (c/  Va2  +  52). 
Square  (3)  and  add  (5), 

r  =  Va2  +  &2  _,.  c2.  (7) 

103.  Solve  for  x  and  y  the  system 

sin  X  +  sin  2/  =  a,        (1) 
cos  X  +  cos  2/  =  6.         (2) 

By  §  40  we  obtain  from  system  (a)  the  equivalent  system  (b) 

2  sin  i  (x  +  ?/)  cos  U^  -  V) 
2  cos  i  (x  +  ?/)  cos  I  (x  -  2/)  =  6, 

Divide  (3)  by  (4),     tan  i  (x  +  2/)  =  a/6.      (5) 

Hence  sin  ^^{x  +  y)  =  ±  a /^a^  +  h\  (6) 

Substituting  the  value  of  sin  \{x-\-y)  in  (3),  we  obtain 

cos  \{x-y)  =  ±  Va2  +  62/2.  (7) 


} 

iral( 
a,         (3)  1 
6.         (4)  J 


(a) 


W 


From 

(5), 

From 

(7), 

Hence 

and 

104. 

Solve  the  system 

sin  X 

cosx 

SYSTEMS  OF  EQUATIONS  139 

X  +  y  =  2  tan-i  {a/b). (8)  | 

X  -  y  =  2  cos-i  (±  Va2  +  W/2).    (9)  J       ^^' 
X  =  tan-i  (a/b)  +  cos-i  ( ±  Va2  +  6V2), 
y  =  tan-i  (a/b)  -  cos-i  (±  Va2+^/2). 


-  sin  y  =  a,  ^ 

—  cosy  =  b.  J 


105.  Solve  for  x  and  y  the  system 

X  sin  ^  +  y  sin  ^  =  a,         (1)  ^ 

X  cos  ^  +  y  cos  ^  =  b.        (2)  j  '  ' 

Since  (1)  and  (2)  are  each  linear  algebraic  equations  in  x  and  y, 
system  (a)  is  solved  as  a  linear  algebraic  system. 

106.  Solve  for  r  and  0  the  system 
r  sin  {6  +  A)  =  a, 


i{e-\-B)  =  b.j 


(a) 


By  [7]  and  [8],  from  (a)  we  obtain  the  equivalent  system  (b). 


r  sin  6  cos  A  +  r  cos  d  sin  ^  =  a,  i 

'  J.  (b) 

r  cos  ^  cos  J5  —  r  sin  ^  sin  B  =b.  J 

Solve  (b)  as  an  algebraic  linear  system  in  r  sin  6  and  r  cos  6  as  the 
unknowns. 

The  resulting  system  can  then  be  solved  for  r  and  6  as  in  example  101. 

107.    Solve  the  system 

cos  (X  +  y)  +  cos  {x-y)  =  2,        (1)  ^ 
sin(x/2)  +  sin(y/2)  =  0,         (2)  J 
for  values  of  x  and  y  less  than  2  tt. 

By  [8]  and  [10],  from  (1)  we  obtain  the  equivalent  equation 
cos  X  cos  y  —  sin  x  sin  y  +  cos  x  cos  y  +  sin  x  sin  y  =  2, 
or  cosxcosy=:l.  (3) 

From  (3),  cos  x  and  cos  y  are  both  + 1  or  both  —1.     Why  ? 

Hence  x  and  y  are  both  coterminal  with  0  or  both  coterminal  with  7t. 


(a) 


(2)/ 


140  PLANE  TRIGONOMETRY 

The  solution  x  =  0,  2/  =  0  of  (3)  satisfies  (2) ;  also  the  solution  x  —  Tt, 
y  =  —  7t,  or  X  =  —  7t,  y  =  7t  oi  (S)  satisfies  (2). 

Observe  that  either  z  =  7t,  y  =  tt,  or  x  =  —  it,  y  =  —  tc  is  &  solution  of 
(3),  but  neither  is  a  solution  of  (2). 

108.  Solve  for  B  and  F  the  system 

W- Fsinh- Rcosh  =  0,^ 
W-  Fcosh-  Rsinh  =  0.) 
Observe  that  this  system  is  algebraic  and  linear  in  R  and  F. 

109.  Eliminate  6  from  the  system 

cc  =  r(^ -sin  ^),         (1) 

y  =  r{l  -cose).         (2) 
From  (2),  y  =  r  vers  d,       or      6  =  vers-i(?//r).  (3) 

From  (2),  cos  d  =  (r  —  y)/r.   .-.  sin  ^  =  ±  ■V2ry  -y^/r.  (4) 

From  (1),  (3),  (4),  x  =  r  vers-i  {y/r)T  V2  ry  -  y^.  (5) 

110.  Eliminate  d  from  the  system 

a  cos  6  -\-bsm6  =  c,         (1)  "i 

dcosd  -\-  esme=f.         (2)  J 

Solving  the  system  for  cos  6  and  sin  6,  we  obtain 

.    n      af-cd  .      ce  —  bf  .„. 

sm  0  =  — ,        cos  ^  = -^-  (3) 

ae  —  bd  ae  —  bd 

Squaring  the  members  of  equations  (3)  and  adding,  we  obtain 

{ae  -  bdf 
...  (ae  -  bdf  =  {af  -  cdf  +  (ce  -  bff. 

111.  Eliminate  6  from  the  system 

acosd  -\-b  sin  ^  =  c,       bcosS  —  a  sin  8  =  d. 

112.  Solve  the  cubic  equation      x^  -Spx  +  q  =  0.  (1) 
Putting  x  —  z/n,  we  obtain 

z3  _  2>-pn^z  +  gn3  =  0.  (2) 

Now  by  (5)  in  example  53,  we  have  the  identity 
cos3^  =  4  cos3^  -  3cos^, 
pr  cos3^-(3/4)cos^ -cos(3^)/4  =  0.  (3) 


CUBIC  EQUATIONS  141 

Comparing  identity  (3)  with  equation  (2),  we  see  that  cos  A  is  a,  root  of 
(2)  when  n  and  A  satisfy  the  conditions 

3pn2  =  3/4,  and  qn^  =  - cos {S  A)  / 4. 
Hence  ?i  =  l/(2Vp), 

and  cosSA  =- 4:qn^  z=-q/{2p^^^).  (4) 

Observe  that  (4)  can  always  be  solved  when  p  is  positive  and  q/{2p^^^) 
is  arithmetically  equal  to  or  less  than  1. 

If  Ai  is  the  principal  value  of  A  which  satisfies  (4),  then  the  values 
Ai  +  27t/S  and  ^i  +  4  ;r/3  also  satisfy  it. 

Hence  the  roots  of  equation  (1)  are 

cos  Ai/n,  cos  (^1  +  2  7t/S)  /n,  and  cos  (^i  +  4  7t/S)  /n, 
i.e.         2  Vp  cos  J-i,  2  VP  cos  (^i  +  2  7r/3),  and  2  VP  cos  {Ai  +  4  tc/S). 

By  Algebra,  we  know  that  the  general  cubic  equation 
z/3  +  3  a?/2  +  6y  +  c  =  0 
can  be  transformed  into  one  of  the  type  (1)  by  putting  y  =  x  —  a. 

113.  Solve  '  a;3  +  6x2  +  9x  +  3  =  0. 
Putting  X  =  ?/  —  2,  we  obtain  y*  —  3  ?/  +  1  =  0. 
Putting  y  =  z/n,  we  obtain  z^  —  S n^z  +  n^  =  0. 
Now         cos8^  -(3/4)cos^  -(1/4) cos 3^  =  0. 

Hence  z  =  cos  A,  when  n^  =  1  /4  and  n^  =  —  cos  3  J.  /4, 

i.e.  when  n  =  l/2,  and  cos 3^  =-1/2  =  cos  120°. 

.-.3^1  =  120°,  or  ^1  =  40° 
Hence  z  =  cos  40°,  cos  (40°  +  120°),  or  cos  (40°  +  240°). 

.-.  y  =  2  cos  40°,  2  cos  160°,  or  2  cos  280°. 
.-.  X  =  -  2  +  2  cos  40°,  -2  -2  cos  20°,  or  -2+2  cos  80°. 

Having  given  sin  15°  =  ( V3  -  l)/(2  V2),  and  cos  15°  =  ( V3  +  l)/(2  V2), 
solve  each  of  the  following  equations  : 

114.  a^ -24a; -32  =  0.  116.    2x3_3a;_i  =  o. 

115.  x3-6x2  +  6x  +  8  =  0.  117.   x3  +  3x2-l  =  0. 

118.    x3  +  4x2  +  2x-l  =0. 


142  PLANE  TRIGONOMETRY 

EXERCISE  XXXm 
Triangles 

1.  Two  towers  are  3  mi.  apart  on  a  plain.  The  angle  of  depression 
of  one,  from  a  balloon  directly  above  the  other,  is  observed  to  be  8°  15'. 
How  high  is  the  balloon  ? 

2.  The  shadow  of  a  tree  101.3  ft.  high  is  found  to  be  131.5  ft.  long. 
Find  the  elevation  of  the  sun. 

3.  A  rock  on  the  bank  of  a  river  is  130  ft.  above  the  water  level.  From 
a  point  just  opposite  the  rock  on  the  other  bank  of  the  river  the  angle  of 
elevation  of  the  rock  is  14°  30'  2V'.     Find  the  width  of  the  river. 

4.  A  rope  38  ft.  long,  when  fastened  to  the  top  of  a  tree  29  ft.  high, 
just  reaches  a  point  in  the  plane  of  the  foot  of  the  tree.  Find  the  angle 
which  the  rope  makes  with  the  ground. 

5.  A  window  in  a  house  is  24  ft,  from  the  ground.  Find  the  inclina- 
tion of  a  ladder  placed  8  ft.  from  the  side  of  the  building  and  reaching 
the  window. 

6.  A  ladder  40  ft.  long  reaches  a  window  33  ft.  high,  on  one  side  of  a 
street.  Its  foot  being  at  the  same  point,  it  will  reach  a  window  21  ft. 
high  on  the  opposite  side  of  the  street.     Find  the  width  of  the  street. 

7.  A  lighthouse  54  ft.  high  is  situated  on  a  rock.  The  angle  of  eleva- 
tion of  the  top  of  the  lighthouse,  as  observed  from  a  ship,  is  4°  52',  and 
the  angle  of  elevation  of  the  top  of  the  rock  is  4°  2'.  Find  the  height  of 
the  rock  and  its  distance  from  the  ship. 

8.  A  man  standing  south  of  a  tower,  on  the  same  horizontal  plane, 
observes  its  angle  of  elevation  to  be  54°  16' ;  he  goes  east  100  yd. ,  and 
then  finds  its  angle  of  elevation  to  be  50°  8'.    Find  the  height  of  the  tower. 

9.  A  pole  is  fixed  on  the  top  of  a  mound,  and  the  angles  of  elevation 
of  the  top  and  the  bottom  of  the  pole  are  60°  and  30°  respectively. 
Prove  that  the  length  of  the  pole  is  twice  the  height  of  the  mound. 

10.  Given  that  the  radius  of  the  earth  is  3963  mi.,  and  that  it  sub- 
tends an  angle  of  57'  2"  at  the  moon.  Find  the  distance  of  the  moon 
from  the  earth. 


TRIANGLES  X43 

11.  Given  that  the  radius  of  the  earth  is  3963  mi.,  and  that  it  sub- 
tends an  angle  of  9"  at  the  sun.  Find  the  distance  of  the  sun  from  the 
earth. 

12.  Solve  example  1  in  §  57  by  the  principles  of  right  triangles. 
The  given  parts  are  a  side  and  two  angles. 

In  fig.  38  draw  AH  ±  BC. 

In  the  right  triangle  AHB,  compute  the  sides  c  and  BR. 

Then  compute  c  in  the  right  triangle  HCA. 

13.  Solve  the  first  four  examples  in  Exercise  XIX  by  the  principles  of 
right  triangles. 

14.  Solve  example  1  in  §  58  by  the  principles  of  right  triangles. 
The  given  parts  are  two  sides  and  an  angle  opposite  one  of  them. 
In  fig.  39  draw  CH  ±  AB. 

Compute  the  sides  CH  and  AH  in  A  ABH. 
Then  compute  B  and  HB  in  A  HBC. 

15.  Solve  examples  1,  3,  5,  and  7  in  Exercise  XX  by  the  principles  of 
right  triangles. 

16.  Solve  the  example  in  §  59  by  the  principles  of  right  triangles. 
The  given  parts  are  two  sides  and  their  included  angle. 

In  A  ABC  draw  BH  ±  CA. 
Compute  CH  and  BH  in  A  CHB. 
Compute  A  and  c  in  triangle  BHA. 

17.  Solve  the  first  four  examples  in  Exercise  XXI  by  the  principles 
of  right  triangles. 

18.  Solve  example  6  in  §  60  by  the  principles  of  right  triangles. 
The  given  parts  are  the  three  sides. 

In  the  A  ABC  draw  AH ±  BC  and  let  x  =  HC. 

Then  b^  -  x'^  =  AH^  =  c^  -  {a  -  x)\ 

Hence  h^  -  x^  =  c^  -  a^  -  x^  i- 2  ax. 

.:  cc  =  (a2  +  62_c2)/(2a). 

Whence  HC  and  b  are  known  in  A  HAC,  and  c  and  BH  in  A  BAH. 

19.  Solve  the  first  four  examples  in  Exercise  XXII  by  the  principles 
of  right  triangles. 


144  PLANE  TRIGONOMETRY 

20.  A  tree  stands  at  a  distance  from  a  straight  road  and  between  two 
milestones.  At  one  milestone  the  line  to  the  tree  is  observed  to  make 
an  angle  of  25°  15'  with  the  road,  and  at  the  other  an  angle  of  45°  17'. 
Find  the  distance  of  the  tree  from  the  road. 

21.  From  the  decks  of  two  ships  at  C  and  D,  880  yd.  apart,  a  cloud  A, 
in  the  same  vertical  plane  as  C  and  D  and  between  them,  is  observed.  Its 
angle  of  elevation  at  C  is  found  to  be  35°,  and  at  D  64°.  Find  the  height 
of  the  cloud  above  the  surface  of  the  sea,  the  height  of  the  eye  in  each 
case  being  21  ft. 

22.  To  determine  the  distance  between  two  ships  at  sea,  an  observer 
noted  the  interval  between  the  flash  and  report  of  a  gun  fired  on  board 
each  ship,  and  measured  the  angle  which  the  two  ships  subtended.  The 
intervals  were  4  seconds  and  6  seconds  respectively,  and  the  angle  48°  42'. 
Find  the  distance  between  the  ships,  the  velocity  of  sound  being  1142  ft. 
per  second. 

23.  In  order  to  find  the  breadth  of  a  river  a  base  line  of  600  yd,  was 
measured  in  a  straight  line  close  to  one  side  of  it,  and  at  each  extremity 
of  the  base  the  angle  subtended  by  the  other  end  and  a  tree  upon  the 
opposite  bank  was  measured.  These  angles  were  53°  and  79°  12'  respec- 
tively.    Find  the  breadth  of  the  river. 

24.  A  straight  road  leads  from  a  town  ^  to  a  town  5,  12  mi.  distant ; 
another  road,  making  an  angle  of  77°  with  the  first,  goes  from  ^  to  a 
town  C,  7  mi.  distant.     Find  the  distance  between  the  towns  B  and  C. 

25.  Two  lighthouses  A  and  ^  are  11  mi.  apart.  A  ship  C  is  observed 
from  them  to  make  the  angles  BAC  =  31°  13'  31"  and  ABC  =  21° 46' 8". 
Find  the  distance  of  the  ship  from  A. 

26.  Two  posts  A  and  B  are  separated  by  a  swamp.  To  find  the  dis- 
tance between  them  a  point  C  is  so  taken  that  both  posts  are  visible  from 
it.  By  measurement,  AC  =  1272.5  ft.,  BC  =  2012.4  ft.,  and  ZACB 
=  41°  9'  11".     Find  the  distance  AB. 

27.  Two  buoys  A  and  B  are  one  half  mile  apart.  Find  the  distance 
from  ^  to  a  point  C  on  the  shore  if  the  angles  ABC  and  BAC  are  77°  7' 
and  67°  17'  respectively. 

28.  The  elevation  of  the  top  of  a  spire  at  one  station.  A,  was  23°  50'  15", 
and  the  horizontal  angle  at  this  station  between  the  spire  and  another 


PROBLEMS  145 

station,  B,  was  93°  4'  W.  The  horizontal  angle  at  B  was  54°  28'  30", 
and  the  distance  between  the  stations  416  ft.  Find  the  height  of  the 
spire. 

29.  In  order  to  find  the  distance  of  a  battery  at  B  from  a  fort  at  F, 
distances  BA  and  AC  were  measured  to  points  A  and  C  from  which 
both  the  fort  and  the  battery  were  visible,  the  former  distance  being 
2000  and  the  latter  3000  yd.  The  following  angles  were  then  measured  : 
Z  BAF  =  34°  10',  ZFAC=  74°  42',  and  Z  FCA  =  80°  10'.  Find  the 
distance  of  the  fort  from  the  battery. 

30.  The  distances  of  two  islands  from  a  buoy  are  3  and  4  mi.  respec- 
tively. The  islands  are  2  mi.  apart.  Find  the  angle  subtended  by  the 
islands  at  the  buoy. 

31.  Two  rocks  in  a  bay  are  c  yd.  apart,  and  from  the  top  of  a  cliff  in 
the  same  vertical  plane  with  the  rocks  their  respective  angles  of  depression 
are  A  and  3  A.     Show  that  the  height  of  the  cliff  is  c  sin  3  J./ (2  cos  A). 

32.  A  person  wishes  to  find  the  distance  between  two  places  A  and  B 
on  opposite  sides  of  a  brook.  He  walks  from  E  to  a  bridge  2  mi.  away. 
Crossing  this  he  continues  his  walk  6  mi.  in  the  same  direction  to  C, 
which  he  knows  to  be  3  mi.  from  A.  If  .4  is  4  mi.  from  the  bridge, 
show  that  AB  =  5.86  mi.,  nearly. 

33.  A  person  at  the  top  of  a  mountain  observes  the  angle  of  depression 
of  an  object  in  the  horizontal  plane  beneath  to  be  45°;  turning  through  an 
angle  of  30°  he  finds  the  depression  of  another  object  in  the  plane  to  be 
30°.  Show  that  the  distance  between  the  objects  is  equal  to  the  height  of 
the  mountain. 

34.  From  a  window  on  a  level  with  the  bottom  of  a  steeple  the  angle  of 
elevation  of  the  top  of  the  steeple  was  40°.  At  another  window  18  ft. 
vertically  above  the  former,  the  angle  of  elevation  was  37°  30'.  Find  the 
height  of  the  steeple. 

35.  Find  what  angle  a  tower  will  subtend  at  a  distance  equal  to  six 
times  the  height  of  the  tower.  Find  where  an  observer  must  station 
himself  that  the  angle  of  elevation  may  be  double  the  former  angle. 

36.  Two  ships  are  a  mile  apart.  The  angular  distance  of  the  first  ship 
from  a  fort  on  the  shore,  as  observed  from  the  second  ship,  is  35°  14'  10"; 
the  angular  distance  of  the  second  ship  from  the  fort,  observed  from  the 
first  ship,  is  42°  11'  53".  Find  the  distance  in  feet  from  each  ship  to 
the  fort. 


146  PLANE  TKIGONOMETRY 

37.  The  sides  of  a  triangle  are  17,  21,  28.  Prove  that  the  length  of  a 
line  bisecting  the  greatest  side  and  drawn  from  the  vertex  of  the  opposite 
angle  is  13. 

38.  Along  the  bank  of  a  river  is  drawn  a  base  line  of  500  ft.  The 
angular  distance  of  one  end  of  this  line  from  an  object  on  the  opposite 
side  of  the  river,  as  observed  from  the  other  end  of  the  line,  is  53° ;  the 
angular  distance  of  the  second  extremity  from  the  same  object,  observed 
from  the  first  extremity,  is  79°  12'.     Find  the  breadth  of  the  river. 

39.  Two  forces,  one  of  410  lb.  and  the  other  of  320  lb.,  make  an  angle 
of  51°  37'.     Find  the  size  and  direction  of  their  resultant. 

40.  An  unknown  force  combined  with  one  of  128  lb.  produces  a 
resultant  of  200  lb.,  and  this  resultant  makes  an  angle  of  18°  24'  with  the 
known  force.     Find  the  size  and  direction  of  the  unknown  force. 


Areas  and  Regular  Polygons 

41.  Two  sides  of  a  parallelogram  are  59.8  ch.  and  37.05  ch.,  and  the 
included  angle  is  72°  10'.     Find  the  area. 

42.  The  three  sides  of  a  triangle  are  49  ch.,  50.25  ch.,  and  25.69  ch. 
Find  the  area. 

43.  One  side  of  a  regular  pentagon  is  25.     Find  the  area. 

44.  One  side  of  a  regular  decagon  is  46.     Find  the  area. 

45.  In  a  circle  with  a  diameter  of  125  ft.  find  the  area  of  a  sector  with 
an  arc  of  22°. 

46.  In  a  circle  with  a  diameter  of  50  ft.  find  the  area  of  a  segment 
with  an  arc  of  280°. 

47.  A  building  is  37.54  ft.  wide  and  the  slope  of  the  roof  is  43°  36'. 
Find  the  length  of  the  rafters. 

48.  What  angle  at  the  center  of  a  circle  does  a  chord  which  is  4/7  of 
the  radius  subtend  ? 

49.  The  side  of  a  regular  pentagon  is  2.     Find  the  radius  of  the 
inscribed  circle. 

50.  The  side  of  a  regular  decagon  is  23.41  ft.     Find  the  radius  of  the 
inscribed  circle. 


REGULAK  POLYGONS  147 

51.  The  perimeter  of  a  regular  polygon  of  11  sides  is  23.47  ft.     Find 
the  radius  of  the  circumscribed  circle. 

52.  The  perimeter  of  a  regular  heptagon  inscribed  in  a  circle  is  12. 
Find  the  radius  of  the  circle. 

53.  Find  the  perimeter  of  a  regular  decagon  circumscribed  about  a 
unit  circle. 

54.  Find  the  perimeter  of  a  polygon  of  11  sides  inscribed  in  a  unit 
circle. 

55.  The  perimeter  of  an  equilateral  triangle  is  17.2  ft.     Find  the  area 
of  the  inscribed  circle. 


FORMULAS 


{sin  A  CSC  A  =  1. 
cos  A  sec  A  =  1. 
tan  A  Got  A  =  1. 

2.  tan  ^  =  sin  ^ /cos  J. 

3.  cot  ^  =  cos  ^/sinyl. 

4.  sin^^  H-  cos^^  =  1. 

5.  tan^  A  -^1  =  sec^  A . 

6.  cot^^  H-  1  =  csc^^. 

7.  sin  (^  +  j5)  =  sin  ^  cos  5  +  cos  A  sin  5. 

8.  cos  (A  -]-  B)  =  cos  ^  cos  B  —  sin  ^1  sin  B. 
tan  ^  +  tan  B 


Paob 


1  —  tan  A  tan  B 
cot  A  cot  5  —  1 


9.   tan(^  -i-B) 

10.  cot  U  +  5) 

^  ^        cot  5  + cot  ^ 

11.  sin  ( J  —  5)  =  sin  ^  cos  5  —  cos  A  sin  B. 

12.  cos  (^  —  5)  =  cos  ^  cos  5  +  sin  A  sin  5. 

.  „     ,       .  .       „.         tan  ^  —  tan  5 

13.  tan(^  -B)  =  rj— — — -• 

^  ^       1  +  tan  A  tan  B 


14.    cot  (A  -  B) 


cot  ^  cot  J3  +  1 
cot  B  —  cot  A 
148 


31 


53 
66 
57 

55 

56 

67 


15. 
16. 

17. 
18. 
19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 
28. 


FORMULAS 

sin  2^  =  2  sin  ^  cos  A. 

cos  2.4  =  cos^ ^  —  sin^  A  (i) 
=  l-2sinM  (ii) 
=  2cosM-l.       (iii) 

2  tan  A 


tan  2^  = 
cot  2A  = 


sin 


1  -  tan2  A 

cot^  yl  -  1 
2  cot  A 

—  cos  .4 


2 
H-  cos  ^ 


2 
cos  A 


-h  cos  ^ 


'    A  _     11  +  cos  A 
"^^2^\l-cos^ 


sm  C  +  sm  Z)  =  2  sin  — - —  cos  — ^ 

C  -  D 


C  +  D    . 
sm  C  —  sm  D  =  2  cos  — - —  sm 


C  ^-  D 
cos  C  4-  cos  2)  =  2  cos  — - —  cos 


2 
C-  7) 


cos  C  —  cos  Z>  =  —  2  sm  — - —  sm  — - — 


sin  C  +  sin  Z)  _  tan^(C  +  i>) 
sin  C  —  sin  D  ~  tan  -J-  (C  —  D) 

a  h  c 

sin  ^  ~  sin  j5  ~"  sin  C 


149 

Paoe 

57 

57 

58 


68 


59 


60 


61 
75 


150 


PLANE  TRIGONOMETRY 


29.   a2  =  &2_|_c2_2^>ccos^ 
^>2  +  c^  -  a^ 


30.    cos  .4 
31. 


2  be 
a-\-b      tan  ^  {A  +  B) 


a  —  b      tan  -I-  (^  —  B) 
3^.    taii^-  =  ^-p^cot- 


33.    sin 


34.   cos 


35.   tan 


^) 


^  /5(g-a) 

2        \       be 

A_     Us  -b)(s-  c) 
2~\      s(s-a) 


Page 

76 


79 


88 


36.  r  =  V(5  -a)(s-  b)  (s  -  c)/s. 

37.  tan(yi/2)=r/(5-a). 

38.  F=Jcsin^/2. 

39.  F  =  -s/s(s-a)(s-b)(s-c). 

,^     „      ft^sin  A  sin  C 

40.  F  = ^r—. — 

2  sin  5 

41.  TT  radians  =  180°  =  2  right  angles. 

42.  N  =  s/r. 

43.  sin(270°±^)  =  cos(180°±^) 

=  -sin(90°±^)  =  -cos^. 

44.  cos  (270°  ±  J)  =  - sin  (180°  ±^) 

=  -  cos  (90°  ±  ^)  =  ±  sin  A. 

45.  tan  (270°  ±  ^)  =  -  cot  (180°  ±  A) 

=  tan  (90°  ±  ^)  =  q:  cot  ^.     , 


89 

92 

95 
96 

'44 


FORMULAS 


151 


46.  sin3^  =  3sin^ -4sin»^. 

47.  cos  3^  =  4  cos*^  —  3  cos  A. 
3  tan  A  —  tan^^ 


48.    tan 3.4 


1  -3tanM 


Page 


136 


49.  In  the  following  table  any  two  expressions  in  the  same 
line  are  equal  arithmetically.  Whether  they  are  like  or  oppo- 
site in  quality  is  known  by  §  22,  when  it  is  known  in  what 
quadrant  A  is.     See  pages  32,  33. 


8\nA 

tan^ 

1 

V^8ec2y(-1 

sec^ 

1 
sec^ 

1 

sin^ 

^/l-cos2^ 
cos^ 

v^H-tan2^ 

1 

COt^ 

csc^ 

V^C8C2^-1 

cos^ 

"^l-sin^A 
sin  A 

tan^ 

1 
tan^ 

"^l  +  cot'A 

1 
cot^ 

cot  A 

C8C^ 

V^l-C082.4 

cos  A 
cos  A 

1 

tan  A 

^^sec2^-l 

1 
^sec^A-1 

sec  A 
sec  A 

^^0802^-1 

sin^ 

1 

cot  A 

^C8C2^-1 

"^l-cos^A 

cos  A 

1 

sec^ 

^l+COt2^ 

cot^ 

CSC^ 

>^l-8in^A 

1 
sin^ 

Vcsc2^-1 

v^l  +  tan2^ 
tan^ 

CSC  A 

^^l  +  cot*^ 

CSC  A 

Vl-COS2vl 

^sec^A-1 

50.  If  sin  0  =  sin  A,e  =  n7r-\-(-  1)M. 

51.  If  cos  0  =  cos  A,  0  =  2  TiTT  ±  A. 

52.  If  tan^  =  tanyl,  ^  =  n7r  +  ^. 

m  :tn 


53.    tan~^?>i  ±  tan~^7i- =  tan" 


1  q=  mn 


Page 

99 
100 
101 

107 


ANSWERS 


EXERCISE  I 


1.  sin  ^  =  2/5,  csc^  =  5/2,  cos^  =V21/5,  sec^  =  5/V21, 
tan  ^  =  2/V21,  cot  A  =  V21  /2  ;  A=2S°  34'  40^  . 

2.  sin  ^  =  4/5,  esc  ^1  =  5/4,  cos  ^  =  3/5,  sec  A  =  6/3, 
tan^  =  4/3,  cot^  =  3/4;  A  =  63° 8'. 

3.  sin  A  =  V7/4,  esc  ^  =  4/V7,  cos  J.  =  3/4,  sec  ^  =  4/3, 
tan  A  =^7  /3,  cot  ^  =  3/ V7  ;  ^  =  41°  24'  30". 

4.  sin  A  =  V8/3,  esc  A  =  3/V8,  cos  ^  =  1/3,  sec^  =  3, 
tan  ^  =  V8,  cot  A  =  1/ V8  ;  ^  =  70°  32'. 

5.  sin  A  =  1/V17,  esc  ^  =V17,  cos^  =4/V17,  sec  ^  =V17/4, 
tan  ^  =  1/4,  cot  ^  =  4  ;  ^  =  14°  2'  15". 

6.  sin  A  =  4/5,  csc^  =  6/4,  cos^  =  3/6,  sec  A  =  6/3, 
tan  A  =  4/3,  cot  ^  =  3/4  ;  A  =  63°  7'  46". 

7.  sin  ^  =  2/v'29,  esc  A  =  V29/2,  cos  A  =  5/V29,  sec  A  =  V29/5, 
tan^  =  2/5,  cot^  =  6/2;  A  =  21° 48' 6". 

8.  sin  A  =  3/VlO,  esc  A  =  VlO/3,  cos  A  =  1/VlO,  sec  A  =  VIO, 
tan  ^  =  3,  cot  ^  =  1  /3  ;  ^  =  71°  34'. 

9.  sin  A  =  4/6,  esc  A  =  5/4,  cos  .4  =  3/5,  sec  A  =  5/3, 
tan  ^  =  4 /3,  cot  ^  =  3/4  ;  A  =  53°  8'. 

10.  sin  A  =  V7/4,  esc  ^  =  4/V7,  cos  J.  =  3/4,  sec^  =  4/3, 
tan  A  =  V7/3,  cot^  =  3/V7  ;  ^  =  41°  24'  30". 

11.  sin  A  =  2/5,  esc^  =  6/2,  cos^  =V21/6,  sec  ^  =  5/V21, 
tan  A  =  2/V21,  cot  A  =  V21/2  ;  ^  =  23°  35'. 

12.  sin  A  =2/3,  csc^  =  3/2,  cos^  =  V5/3,  sec^  =  3/V5, 
tan  ^  =  2/ V5,  cot  ^  =  V5/2  ;  ^  =  41° 48'  30". 

153 


154  PLANE  TRIGONOMETRY 

13.  sin  JL  =  4/V17,  csc^=V17/4,  cos^  =  1/V17,  8ecA=y/n, 
tan  J[  =  4,  cot  ^  =  1  /4  ;  J.  =  75°  57'  49''. 

14.  sin  ^  =  1  /  V50,  CSC  A  =  V50,  cos  J.  =  7  /  V^O,  sec  A  =  V50/7, 
tan  ^  =  1/7,  cot  ^  =  7  ;  ^  =  8°  7'  48". 

15.  sin  ^  =  9/V82,  esc  4  =  V82/9,  cos  A  =  1/V82,  sec  A  =  V82, 
tan  ^  =  9,  cot  ^  =  1  /9  ;  ^  =  83°  39'  35". 

16-18.    For  answers  see  table  on  page  151. 

EXERCISE  n 

1.  sin  60° ;  cos  30° ;  tan  55° ;  cot  75° ;  esc  5° ;  sec  14° ;  cos  16°  46' ; 
sin  24°  17'. 

2.  45°.  4.    18°.  6.    5°.  8.    90°/(m  +  n). 

3.  30°.  5.   36°.  7.    5°.  9.   60°/(c  -  1). 

EXERCISE  m 

1.  5  =  65°,  6  =  64.335,  c  =  70.98. 

2.  A  =  35°,  a  =  14.281,  c  =  17.434. 

3.  5  =  25°,  a  =  63.441,  6  =  29.582. 

4.  ^  =  75°,  a  =  74.64,  c  =  77.28. 

5.  ^  =  55°,  6  =  35.01,  c  =  61.05. 

6.  ^  =  35°,  6  =  49.152,  a  =  34.416. 
r.  -4  =  20°,  B  =  70°,  c  =  106.4. 

8.  ^  =  25°,  5  =  65°,  c  =  55.15. 

9.  ^  =  20°,  B  =  70°,  a  =  34.2. 

10.  A  =  20°,  B  =  70°,  6  =  46.985. 

11.  A  =  15°,  a  =  10.352,  6  =  38.636. 

12.  B  =  80°,  a  =  5.289,  c  =  30.45. 

13.  5  =  70°,  a  =  27.36,  6  =  75.176. 


ANSWERS  155 

14.  A  =  65°,  b  =  13.989,  c  =  33.09. 

15.  A  =  65°,  B  =  25°,  b  =  14.086. 

16.  ^  =  15°,  ^  =  75°,  c  =  51.75. 

-    EXERCISE   rV 

1.  160.7  yd.  5.    42.68  ft.  9.    81.98  ft. 

2.  50°.  6.    107.22  ft.  11.    1174.6  ft. 

3.  96.06  ft.      7.  136.63  ft.         12.  3770  ft. 

4.  122.02  ft.      8.  44.78  ft.,  37.58  ft.    13.  98.097  ft.,  68.69  ft. 

14.  36.08  ft.,  154.23  ft. 

15.  33.51  ft.,  28.12  ft.  to  nearest  tower. 

16.  74.24  ft.,  51.98  ft.  17.   378.21  ft.,  417.17  ft. 

18.  h  =  altitude  =  32.14  ft.,  c  =  base  =  76.6  ft., 
Q  =  area  =  1231  sq.  ft. 

19.  ZC  =  40°,  ZA  =  70°,  h  =  93.97  ft.,  Q  =  3213.8  sq.  ft. 

20.  61.04  ft.,  ZA  =  35°,  ZC=  110°. 

21.  h  =  62.836  ft.,  r  =  76.71  ft.,  Q  =  2764.8  sq.  ft. 

22.  h  =  107.23  ft.,  r  =  118.3,  Q  =  5361.3  sq.  ft. 

23.  c  =  r=:  57.74  ft.,  Q  =  1443.5  sq.ft.       28.492.4  ft. 

24.  20°.  29.    2515  ft. 

25.  R  =  274.75  ft.,  area  =  237150sq.  ft.      30.    125  (V3  +  3)  ft. 

26.  25°;  14.09  ft.  31.    30°. 

27.  917.136  ft. 

EXERCISE  V 

1.  2d  qdt.  4.    2d  qdt. ;  4th  qdt.  7.    2d  qdt. ;  1st  qdt. 

2.  4th  qdt.  5.    2d  qdt.;  1st  qdt.  8.    4th  qdt. ;  3d  qdt. 

3.  2d  qdt.  6.    4th  qdt. ;  3d  qdt.  9.    3d  qdt. ;  3d  qdt. 


156  PLANE  TRIGONOMETRY 

10.  847°  is  coterminal  with  127°;  1111°  with  31°;  -225°  with  135°; 

-  300°  with  60°  ;  942°  with  222°  ;   -  1174°  with  -  94°  or  266°. 

11.  405°  and  1125°;  -  315°  and  -  675°;  390°  and  750°;  -  330°  and 

-  690° ;  460°  and  820°  ;   -  260°  and  -  620° ;  560°  and  1280°  ; 

-  160°  and  -  520°;  350°  and  710°;  -  370°  and   -  730°;  260° 
and  620° ;   -  460°  and  -  820°. 

12.  -75°;  15°.                                     15.  -  224°  22' 17'';  -134°  22' 17". 

13.  -  138°  ;   -  48°.                              16.  122°  14'  21"  ;  212°  14'  21". 

14.  -205°  17' 14";  -115°  17' 14".    17.  255°  28' 42";  345°  28' 42". 
18.  60°.          19.    175°.          20.   30°.  21.    340°.          22.   317°. 

EXERCISE  VI 

1.  MP  =  -  2,  OP  =3,  OJlf  =  T  V5  ;  CSC ^  =  -3/2,  cos  ^=  T  \/5/3, 
sec  J.  =T  3/V5,  tan  J.  =±  2/V5,  cot^  =±  V5/2. 

2.  3f P  =  ±  5,  OJIf  =  ±  2,  OF  =  V29  ;  cot  ^  =  2/5, 

sin  ^  =  ±  5/V29,  esc  ^  =  ±  V29/5,  cos  ^  =  ±  2/ V29, 
sec^  =±  V29/2. 

3.  JWP  =  ±  3,  Oilf  =  =F  1,  OP  =  VIO ;  cot  JL  =  -  1  /3, 
smA=±  3/VlO,  CSC  ^  =  ±  VlO/3,  cos  ^  =  T  1  /VIO, 
sec  J.  =  ^  VIO. 

4.  OM  =  2,  OP  =  S,  MP  =  ±y/5',  sec  A  =  3/2,  sin  J.  =  ±  V5/3, 
esc  J.  =  ±  3  /  V5,  tan  J.  =  ±  V5/2,  cot  ^  =  ±  2  /  V5. 

5.  csc^  =-  8/7,  cos^  ==F  V15/8,  sec^  =T  8/V15, 
tan^  =  ±  7/V15,  cot  J.  =:±  V15/7. 

6.  cot^  =  1/7,  sin  J.  =±  7/(5  V2),  csc^  =±  5  V2/7, 
cos  ^  =  ±  1  /(5  V2),  sec  J.  =  ±  5  V2. 

7.  sec  A  =-7/3,  sin  ^  =±2  VlO/7,  csc^  =±7/(2  VIO), 
tan^  =T  2  VlO/3,  cot^  =T  3/(2  VIO). 

8.  JfP  =  ±3,  OJlf=±  5,  OP  =  V34;  tan^  =  3/5,  sin  ^  =  ±  3/V34, 
csc^  =  ±  V34/3,  cos  ^  =  ±  5/V34,  sec  ^  =  ±  V34/5. 

9.  sec^  =-  5/4,  sin^  =±  3/5,  esc  ^  =  ±  5/3,  tan  ^  =^  3/4, 
cot^  =T  4/3. 

10.    OM  =1,  OP  =  2,  ifP  =  ±  V3  ;  cos  ^  =  1/2,  sin  A=±  V3/2, 
CSC  J.  =  ±  2  /  V3,  tan  A=±  V3,  cot  ^  =  ±  1  /  V3. 


ANSWERS  167 

11.  cos  ^  =  -  2/3,  sin  ^  =  ±  V5/3,  esc  ^  =  ±  3/ V5, 

tan^  =qF  V5/2,  cot^  =^2/y/b. 

12.  sin  A  =-3/5,  cos^  =t4/6,  sec4  =:f  5/4,  tan  ^  =±3/4, 
cot  J.  =±4/3. 

EXERCISE  Vn 


1.  csc^  =-3/2,  cos^  =T^1  -4/9  =  T  V5/3,  sec^  =  rp  3/ V5, 
tan  J.  =±  2/V5,  cot^  =±  V5/2. 

2.  sec  ^  =  3,  sin^  =  ±  2  V2/3,  csc^  =  ±  3/(2  V2),  tan^  =  ±  2  V2, 
cot  J.  =±  1/(2  V2). 

3.  CSC  ^  =  5,  cos  ^  =  ±  2  V6/5,  sec  ^  =  ±  5/(2  V6), 
tan  ^  =  ±  1  /(2  V6),  cot  ^  =  ±  2  y/Q. 

4.  sec  J.  =-  4/3,  sin^  =±  V7/4,  esc  ^  =±  4/V7, 
tan^  ==F  V7/3,  cot^  =T  3/V7. 

5.  cot  J.  =-  3/4,  sin^  =±  4/5,  esc  ^  =  ±  5/4,  Cos^  =q:3/5, 
sec  A  =  T  5/3. 

6.  tan  J.  =  -  1/2,  sin  ^  =  ±  1/ V5,  esc  ^  =  ±  y/b,  cos  J.  =  =F  2/ V5, 
sec^  =T  V5/2. 

7.  tan^  =  2/3,  sin  ^  =±2/V13,  esc  ^  =±V13/2, 
cos^  =±3/V13,  sec^  =±V13/3. 

8.  cot^  =  2/5,  sin^  =±  5/V29,  esc  ^  =±  V29/5, 
eos^  =±2/V29,  see^  =±V29/2. 

9.  sin  ^  =-  1/V3,  eos^  =T\/(2/3),  sec^  =T  V(3/2), 
tan  ^  =  ±  1  /  V2,  cot  J.  =  ±  V2. 

10.  cos^  =  1/4,  sin^  =  ±V15/4,  esc  J.  =±4/V15,  tan^  =±  V15, 
cot^  =±  1/V15- 

11.  cot^  =_i/V7,  sin  A=±V14/4,  ese^=±4/V14, 
eos^  =±  V2/4,  see^  =t4/V2. 


12.    sec  A  =  —t  sin  ^  =  ± '— ,  esc  A  =± 


m  c  Vc^ 

Vc2  —  m2  ^  m 

tan  A=± ,  cot  J.  =  ± 


m  Vc2 

13-17.    See  table  on  page  151. 


158  PLANE  TRIGONOMETRY 

EXERCISE  IX 

1.  sin  12°.  5.    -cos 25°  9.   cos 30°.  13.   esc 36°. 

2.  -tan  43°.      6.    sin  6°.  10.    cot  25°.  14.    -tan  16°  54'. 

3.  sin  17°.  7.    -cot  24°  11.    -cot  26°.        15.    -tan  33°  39'. 

4.  cos  24°.  8.   sin  22°.  12.    -esc  23°. 

EXERCISE  X 

1.  The  cosine  of  the  sum\  _  j      cos  first  ■  cos  second 

of  any  two  angles     J  ~  \  —  sin  first  •  sin  second. 

2.  (V2+V6)/4.  6.  1;  0. 

3.  (V6-V2)/4.  7.  0;  1. 

4.  (V6-V2)/4;  (V6  +  V2)/4.  8.  ^'^(1  +  V42);  tV(V21 -4  V2). 

5.  (V6-V2)/4;  (V6  +  V2)/4.  9.  ^V(8  +  5V3);  ^^(2^6-^16). 

EXERCISE  XI 

1.  The  cosine  of  the  difference')  _  f      cos  first  -  cos  second 

of  any  two  angles  J  ~  I  +  s,m  first  •  sin  second. 

2.  (V6-V2)/4;  (V6  +  V2)/4.       3.    (V6-V2)/4;  (V6  +  V2)/4. 

4.  (2  V2 -V15)/12;  (2V30  +  1)/12. 

5.  (3V5-2V7)/12;  (6  +  V35)/12. 

EXERCISE   Xn 

1.  The  tangent  of  the  difference  1  _  f  the  difference  of  their  tangents 

of  any  two  angles  /  ~~  1  1  +  product  of  their  tangents 

2.  (V3  +  1)/(V3-1).  4.    1;  7. 

3.  (V3-1)/(V3  +  1).  5.    1;  1/7. 

,_    ^      /  ^    .   TJX        cot^  +  cot^    .^      ..       _.         cot  B- cot ^ 
12.   tan  {A  +  B)  = ;  tan  {A  —  B)  = 


cot  J.  •  cot  J5  -  1  cot  ^  •  cot  J5  +  1 

-.  o        ^  /  >•    .   T>x      1  —  tan  ^  •  tan  J5  .      ^ ,  .       _,      1  +  tan  ^  •  tan  B 

13.   cot  (A  +  B)= -—  ;  cot  (A-B)  =  —^ 

tan  ^  +  tan  J5  tan  -4  -  tan  B 


ANSWERS  169 

EXERCISE  Xm 

-     ^,     ,  4.    X  *    .  ,        twice  the  tangent  of  the  angle 

1.  The  tangent  of  twice  an  angle  = — 

1  —  (the  tangent  of  the  angle)^ 

2.  V3/2;  1/2;   V3. 

3.  V3/2;   -  1/2;   -V3. 

4.  2  sin  3^  cos  3  ^  ;  cos2  3  ^  -  sin2  3  ^,  1  -  2  sin^  3  A,  or 
2cos2  3^  -  1;  2  tan  3^ /(I  -  tan2  3^). 

5.  2  sin  (3^/2)  cos  (3^/2);  cos2(3^/2)  -  sin2(3^/2),  or 
l-2sin2(3^/2),  or  2  cos2(3^/2)  -  1 ; 
2tan(3^/2)/[l-tan2(3^/2)]. 


EXERCISE  XIV 

1  11^  T  X    J.  1  +  COS  angle 

1.  cos  half  an  angle  =  square  root  of ^— ; 

tan  TiaJf  an  angle  =  square  root  of  —  • 

1  +  cos  angle 

2.  V2-V2/2;   V2  +V2/2;  ^P  ~  ^^  =^S  -  2  ^2. 

\2  +  V2 

3.  V2-V3/2;   V2TV3/2;   V?  -  4  V3. 

4.  V3/3;   V6/3;   V2/2. 

-        /I  -g       11  +  a       /I  -  a 

_        /I  -  cos  2  ^        /I  +  cos  2  ^        /I  - 
®-    V 2 '  V ~2 •  VlT 


COS  2^ 
1  —  cos  4  ^        /I  +  COS  4  ^4        /I  —  cos  4  J. 


.        /I  -  cos  4  ^        /I  +  cos  4  .4        11  - 

V     2     '  \     2     '  Vrr 

,        /I  -  cos  6  ^        /I  +  cos  6  ^        /I 

^-  \ — ^ — '  V — 2 — '  Vi 


cos  4^ 
1  —  cos  6  J. 


+  cos  6  J. 


160 


PLANE  TRIGONOMETRY 


EXERCISE  XV 


}■  { 


The  difference  of  the  sines  of 
any  two  angles 

The  sum  of  the  cosines  of  any  1  _ 
two  angles  j  ~ 

The  difference  of  the  cosines^  _ 
of  any  two  angles  J  ~ 


twice  cos  half  sum  into 
sin  half  difference. 

J  twice  cos  half  sum  into 
\     cos  half  difference. 

r  twice   sin  half  sum   into 
\     sin  half  difference. 


15.    (1)  sin  {A-]-  B)  =  (2  V2  +  V3)/6  : 
cos(^  +  J5)  =  (2V6-l)/6; 
sin  2  A  =:V3/2; 
cos  2^  =  1/2; 


sin  {A-B)z=  (2  V2  -  V3)/6 ; 
cos  (J.  -  1?)  =  (2  V6  +  l)/6 ; 

sin2E  =  4  V2/9; 

cos2JS  =  7/9; 


(2)  sin  ( J.  +  B)  3z  -  (2  V2  -  V3)/6 ;  sin  (^  -  ^)  =  -  (2  V2  +  V3)/6 ; 
cos  U  +  -B)  =  -(2  V6  +  l)/6 ;  cos  {A  -  B)=-  (2  V^  -  l)/6  ; 
sm2^  =  V3/2;  sin  2B  =  -  4  V2/9; 

cos2^  =  l/2;  cos2B  =  7/9. 


16.    (1) 


(2) 


tan  {A-\-  B)  : 

tan  {A  -  B) 

cot  (^  +  B) 

cot  U  -  -S) 

sec  {A  +  B) 

esc  ( J.  +  ^) 

tan  2  4 

cot  2  ^ 

sec  2B 

CSC  2  B 

tan  (^  +  B) 

tan  (^  -  ^) 

cot  {A  +  B) 

cot  (^  -  B) ; 

sec  (^  +  B) 

CSC  (A  +  B) 

tan  2  ^ 

cot  2^ 

sec  2  B 

CSC  2  B 


(2V2+V3)/(2V6-1): 

(2  V2  -  V3)/(2  V6  +  1) 

(2V6-1)/(2V2+V3) 

(2V6  +  1)/(2V2-V3) 

6/(2  V6-1); 

6/(2V2  +  V3); 

V3; 

1/V3; 

9/7; 

9/(4  V2). 

(2V2-V3)/(2V6  +  1) 
(2V2+V3)/(2V6-1) 
(2V6  +  1)/(2V2-V3) 
(2V6-1)/(2V2+V3) 
-6/(2V6  +  l); 

-6/(2V2-V3); 

V3; 

1/V3; 

9/7; 

-9/4  V2. 


ANSWERS  161 


EXERCISE  XVI 

9.   sin(3a;/4)  =  ±V[l  -  cos(3x/2)]/2  ; 
cos(3x/4)=±V[l  +  cos(3x/2)]/2; 


tan(3x/4)=±Vl  -  cos(3x/2)/ Vl  +  cos(3x/2). 

10.    sin  (3x/4)  =  2sin(3x/8)cos(3x/8); 
cos  (3x/4)  =  cos2(3x/8)  -  sin2(3ic/8) ; 
tan(3x/4)  =  2taii(3x/8)/[l  -  tan2(3x/8)]. 

16.  (1)  sin  {A  +  B)  =  {2+^l5)/6;        sin  {A  -  B)  =  {2  -  V15)  /6  ; 

cos(^+  B)  =  {^/6  -  2  V3)/6;  cos(^  -  B)  =  (V5  +  2  V3)/6 
sin2^=4V5/9;  cos2^  =  l/9; 

sin2J5=V3/2;  cos25  =  -l/2. 

tan2^=:4V5;  cot  2^  =  •v/5/20  ; 

tan2B  =  -V3;  cot2B  =  -V3/3. 


EXERCISE   XVn 

1.^=23%  6  =  11.779,  c=  12.796. 

2.  B  =  52°,  b  =  10.355,  c  =  13.14. 

3.  J5  =  75°,  6  =  6.7614,  a  =  1.8117. 
A.   A=  40°,  a  =  16.782,  c  =  26.108. 

5.  A=  34°  22'  9'',  B  =  55°  37'  51",  6  =  0.51176. 

6.  ^  =  33°  8'  56'',  B  =  56°  51'  4",  c  =  499.26. 
1.  A=  39°  49'  22",  B  =  50°  10'  38",  a  =  48.863. 
S.  B  =  81°,  a  =  148.41,  c  =  948.68. 
9.  ^  =  49°  53'  53",  B  =  40°  6'  7',  c  =  4.4632. 

10.  5  =  43°  37',  a  =  3821.5,  6  =  3641.3. 

11.  ^  =  35°  53' 56". 5,  5  =  54°6'3".5,  6=731.23. 

12.  A  =  66°  51',  a  =  176.53,  c  =  191.99. 


162  PLANE  TRIGONOMETRY 

13.  A  =  71°  22',  a  =  2.4099,  b  =  .81268. 

14.  B  =  58°  15',  b  =  77.632,  c  =  91.294. 

15.  ^  =  32°  10' 15'',  J5  =  57°  49' 45",  a  =  388.45. 

16.  -4  =  7°  53' 42",  6  =  644.11,  c  =  650.27. 

EXERCISE  XVin 

1.  C=180°-2^,  r  =  c/ {2 -cos A),  h  =  c-tSinA/2. 

2.  J.  =  90°  -  0/2,  r  =  ^/cos(C/2),  c  =  2  ^  •  tan(C/2). 

3.  ^  =  tan-i(2/i/c),       C  =  2  •  tan-i(c/2/i),  r  =  V4¥T^/2. 

4.  r  =  2.055,       71=1.6853,        ^  =  55°  5' 30",         Q  =  1.9819. 

5.  r  =  7.706,        c  =  3.6676,         C  =  27°  32',  Q=  13.7253. 

6.  Let  X  =  length  of  rafter  and  Q  =  area  of  roof ;   then,  since  the 
roof  projects  one  foot  over  the  side  of  the  barn,  we  have  : 

a;  =  21 /sin  45°,     Q  =  2-x-82;    x  =  29.698,     Q  =  4870.44. 

7.  r  =  1.61804,  7i  =  1.53882,  F=  7.69417. 

8.  r  =  11.2692,  71  =  10.8852,  2?^=  380.99. 

9.  r  =  1.0824,  c  =  8284,  F=  3.3136. 

10.  r  =  1.5994,  7i  =  1.441,  p  =  9.715. 

11.  h  =  28.971,  r  =  31.357,  A  =  (difference  between  polygon  and  in- 
scribed circle)  =  144.51,  Be  =  307.8. 

12.  A  =14.536,        r  =  16.134,         A  =  48.48,        Dc  =  105.47. 
15.    99.64  sq.  ft.  16.   2.393. 

EXERCISE   XIX 

1.  ^  =  65°  15',  6  =  95.6025,  c  =  89.648. 

2.  B  =  72°14',  a  =  75.132,  c  =  92.788. 

3.  ^  =  31°  20',  6  =  184.896,  c  =  191.978. 

4.  C  =  60°,  a  =  255.38,  6  =  282.56. 

5.  ^  =  15°  43',  6  =  222.1,  c  =  321.08. 


ANSWERS  163 

6.  J5  =  66°,  a  =  765.43,  c  =  1035.4. 

7.  1253.2  ft.  9.    300  yd.  11.    294.77  ft. 

8.  1116.6  ft.  10.    12296  ft.,  13055  ft.  12.    4211.8  ft. 

EXERCISE   XX 

1.  ^  =  32°  25' 36",  C=  106°  24' 24'',  c  =  259.4. 

2.  A=  28°  20'  48",  C  =  39°  35'  12",  a  =  293.56. 

3.  J5  =  32°36'9",  C  =  83°  33' 5",  c  =  6.621. 

4.  Bi  =  51°  18'  22",  Ci  =  88°  41'  38",  Cx  =  218.525  ; 
Bi  =  128°  41'  38",  Ca  =  11°  18'  22",  Cg  =  42.853. 

5.  ^1  =  31°  57'  46",  Ai  =  120°  44'  14",  ai  =  120.31  ; 
B2  =  148°  2'  14",  A2  =  4°  39'  46",  03  =  11.379. 

6.  Impossible. 

7.  Ci  =  46°  18' 40",  ^1  =  93°  9' 13",  ai  =  69.4567  ; 
C2  =  133°  41'  20",  A2  =  5°  46'  33",  ag  =  7.0005. 

8.  ^1  =  51°  18' 27",  d  =  98°  21' 33",  Ci  =  43.098  ; 
A2  =  128°  41'  33",  C2  =  20°  58'  27",  d  =  15.593. 

9.  ul  =  54°  31' 13",  C  =  47°44'7",  c  =  50.481. 

10.  ^1  =  24°  57' 54",  Ci  =  133°  47' 41",  Ci  =  616.67  ; 
^2  =  155°  2' 6",  C2  =  3°43'29",  C2  =  55.41. 

11.  ^1  =  16°  43' 13",  ^1  =  147°  27' 47",  ai  =  35.519; 
B2  =  163°  16'  47",  Ai  =  0°  54'  13",  Oa  =  1.0415. 

EXERCISE   XXI 

1.  ^  =  42°  50' 58",  5  =  64°  9' 2",  c  =  374.06. 

2.  5  =  132°  18' 28",  C=  14°  34' 23",  a  =  67.75. 

3.  A  =  109°  15'  30",  B  =  45°  4'  30",  c  =  440.45. 

4.  ^  =  60°  44'  39",  B  =  47°  21'  21",  c  =  966.28. 

5.  B  =  54°  1'  13",  C  =  63°  49'  9",  a  =  44.825. 

6.  902.94  ft.  8.  1331.2  ft.  10.  10.532  mi. 

7.  13.27  mi.  9.  4.8112  mi.  11.  9.6268  mi. 


164  PLANE  TRIGONOMETRY 


EXERCISE   XXn 


1. 

A  =  74°  40'  18'', 

B  =  47°  46'  38", 

C  =  57°  33'  4". 

2. 

A  =  59°  19'  14", 

B  =  68°  34'  8", 

C  =  52°  6'  40". 

3. 

A  =  45°  11'  50", 

B  =  101°  22'  12", 

C  =  33°  25'  58". 

4. 

A  =  11°  33'  52", 

B  =  49°  8'  3", 

C  =  59°  18'  5". 

5. 

A  =  54°  3'  10", 

B  =  30°  47'  22", 

C  =  95°  9'  24". 

6. 

Bi  =  41°  41'  26", 
Bi  =  138°  18'  34", 

Ci  =  111°  52'  34", 
C2  =  15°  15'  26", 

ci  =  177.2; 
C2  =  50.248. 

7. 

A  =  17°  16'  11", 

B  =  28°  43'  49", 

c  =  14.424.                    1 

8. 

A  =  18°  12'  22", 

B  =  135°  50'  46", 

C  =  25°  56'  52".            ' 

9.    Since  c>a,  C>A> 90°,  which  is  impossible,  as  a , triangle  cannot 
have  two  obtuse  angles.                                                                                 , 

10. 

B  =  48°  34'  44", 

A  =  49°  38'  16", 

a  =  76.015. 

11. 

B  =  145°  35'  24", 

C  =  7°  11'  36", 

a  =  104.57. 

12. 

A  =  57°  52'  44", 

B  =  70°  17'  24", 

C  =  51°  49'  50". 

13. 

Ai  =  70°  12'  48", 
A2  =  109°  47'  12", 

Bi  =  57°  22'  56", 
^2  ==  17°  48'  32", 

61  =  28.79; 

62  =  10.454.                     \ 

14. 

A  =  32°  44'  40", 

C  =  63°  48'  28", 

b  =  137.39.                    ; 

15. 

6=143.52, 

B  =  146°  43'  10", 

C  =  14°  4'  7".                i 

16. 

A  =  67°  55'  15", 

B  =  54°  4'  45", 

c  =  85.36.                      i 

17. 

44°  2'  9",  51°  28'  11' 

,  84°  29' 40". 

18. 

44°  2'  56". 

20.    N.  4°  23'  2" 

W.,  or  S.  4°23'2"  W. 

19. 

60°  51'  8". 

21.    60°. 

EXERCISE  XXm 

2.  1931.8.  3.  44770;  781.617;  149689;  314543. 

4.  123794  ;  53596.3  ;  10.6665  ;  Fi  =  11981.7,  F2  =  2349.63. 

5.  1016.23;  30.858;  1430.3;  170346. 

6.  3891.64;  3319.38;  9229.4;  31246.4. 


ANSWERS  165 

EXERCISE  XXIV 

1.  9;r/4,  17;r/4,  -7  7r/4,  -15;r/4;  13;r/4,  21;r/4,  -3  7r/4, 
ll7r/4;  7;r/2,  11 7r/2,  -  ;r/2,  -  5  7r/2  ;  9  7r/2,  13;r/2,  -3;r/2, 
7  7zr/2;  7;r/3,  13  7r/3,  -  5  7r/3,  -  11 7r/3;  8;r/3,  14  7r/3,  -4;r/3, 
10;r/3;  137r/6,  25;r/6,  -ll7r/6,  -237r/6;  177r/6,  29;r/6, 
7  7r/6,   -19;r/6. 

2.  120°.  4.    900°.  6.    ;r/4.  8.    ;r/2. 

3.  300°.  5.    135°.  7.    3  7r/4.  9.    3  7r/2. 

10.  12859/7560,  or  1.7.  14.  ;r/4,  3;r/4. 

11.  115643/37800,  or  3.06.  15.  -  7r/6,  7r/3. 

12.  35827/47250,  or  0.76.  16.  -  ;r/4,  ;r/4. 

13.  1516273/2268000,  or  0.67.  17.  -  7  ;r/6,  -  2  7r/3. 

18.  sin(7r/6)  =  l/2,  cos{7r/6)  =  V3/2,  tan(;r/6)  =  V3/3. 

19.  sin(7r/4)=V2/2,  cos(7r/4)  =  V2/2,  tan(;r/4)=l. 

20.  sin(7r/3)  =  V3/2,  cos(;r/3)  =  1/2,  tan(;r/3)  =  V3. 

21.  sin(;r/2)  =  1,  cos(;r/2)  =  0,  tan(7r/2)  =  oo. 

22.  sin  TT  =  0,  cos  TT  =  —  1,  tan  tc  =  0. 

23.  157/70  or  2.24;  157  (57°  17' 44^8)/ 70. 

24.  55/98;  55  (57°  17' 44^8)/ 98. 

25.  3/4;  3(57°17'44''.8)/4. 

26.  11/210  in.  29.    3980^^  mi. 

EXERCISE   XXV 

1.  e  =  UTt -\- {-l)n{±7r /2)  =  n7t  ±7t/ 2. 

2.  n7t±7t/i.  6.    n7t±7t/Q.  10.  2n7r±7t/3. 

3.  n7r±7r/4.  7.    2n7t±2  7t/S.  11.  2n7t-\-7t/4. 

4.  n7r±7t/e.  8.   n7r  +  (-1)«  •  7r/6.  12.  n7t±7t/6. 

5.  nit±7t/3.  9.   n;r  +  (-l)«  •  (7r/3).  13.  titt  +  7r/4. 


166  PLANE  TRIGONOMETRY 

14.  nTC  +  ;r/4,  mt  +  7t/3.  16.   utc  ±  ;r/4. 

15.  nje  +  5  7t/6,  uTt -^  2  7r/3.  17.   n  •  180°  +  53°  7'  45''. 

18.  (2  w  +  1) .  180°,  2  n  •  180°  +  53°  8'. 

19.  [n;r +  (-!)« ;r/4]/5.  22.  {2n7t  +  7t/2)/d,  2 mt- 7r/2. 

20.  2n7r/9,  2n;r.  23.   (2n7r  ±  7r/2)/(m  ±  r). 

21.  (nnr  + 7r/2)/(r  +  l).  24.   2w7r  +  ;r/4,  2n7r  -  3  7^/4. 
^  must  be  in  the  first  or  third  quadrant.     Why  ? 

25.  2n7r-;r/6,  2n7r  +  6;r/6,  2w7r-7r/3,  2M;r  +  2;r/3. 
d  must  be  in  the  second  or  fourth  quadrant.     Why  ? 

26.  2n7r- 7r/4,  2n7r  +  3  7r/4  28.   (2n  +  l)7r  +  ;r/4. 

27.  n;r/3  +  7r/6.  29.   2n;r-;r/6. 

30.  7  7r/12  and  ;r/4;    (2  m  +  n)  7r/2  ±  tt/G  +  (- l)«(;r/12)  and 
(n  —  2m);r/2  +  (—  l)«;r/12  T  71^/6,  where  m  and  n  are  any  integers. 

31.  25;r/24     and     19;r/24;    {m  +  2n)  tc /2  +  7t /S  ±  7t /12    and 
(2n  —  m)7t/2±7t/12  —  7i/S,  where  m  and  n  are  any  integers. 

EXERCISE  XXVI 

1.  nn/^,  {2n7t  ±7t/S)/3.  8.  UTt  ±  7t/4,  2n7t  ±27C /S. 

2.  TiTi:  ±  7t / 4:,  2  uTt  ±  7t / S.  9.  2ii7r  +  7r/6  ±  7r/4. 

3.  n;r  ±  ;r/4,  n7r-(-l)»;r/6.  10.  2n;r  +  ;r/4. 

4.  »i7r/2  ±  7r/8,  2n7r/3  ±  7r/9.  11.  2n7t  -  tc/S  ±  7t/4. 

5.  2n;r/3±7r/6,  n;r4-(-l)'»7r/6.  12.  2?i;r  +  7r/4  ±  ;r/5. 

6.  n;r/3,  (2w  +  l/3)7r/4.  13.  2  n  •  180°  +  68°  12'. 

7.  n7t/S,n7C±7t/S.  14.  2w  •  180°±60°-33°4r24". 

EXERCISE  XXVn 

1.  (2-V3)/2;  (2-V2)/2;  1/2;  (2+V2)/2;  0;  1;  2;  1. 

2.  1/2;  (2-V2)/2;  (2-V3)/2;  (2-V2)/2;  1;  0;  1;  2. 


ANSWERS  167 

3.   riTT +(-!)»  •7r/4;  n;r -(-1)«  •  7r/3 ;  2n7r±^/6;  2n;r±2;r/3; 
mt-\-7t/Q\  nit- It /Z;  mt-it/^\  n7t+7t/o. 

13.  1/2.  15.    V3.  17.    V3.  19.    13. 

14.  ±V2/2.         16.    -8,1/4.  18.    ±V21/14. 

EXERCISE  XXIX 

1.  (1/V2  +  i/V2)  •  V2,  or  (cos  ;r/4  +  isin  7r/4)  •  V2. 

2.  [l/V2  +  i(-l/V2)]-  V2,  or  [cos(- ;r/4)  +  isin(- ;r/4)]  •  V2. 

3.  [-l/V2  +  i(-l/V2)]- V2, 

or  [cos(-3;r/4)  +  isin(-3  7r/4)].  y/2. 

4.  (_  1/2  +  i  V3/2)-2,  or  (cos2  7r/3  +  isin  2  ;r/3)  •  2. 

5.  (- V3/2  +  i/2)  -2,  or  (cos5  7r/6  +  isin  5  7r/6)  •  2. 

6.  [- V3/2  +  i(-  1/2)]  .2,  or  [cos(- 5  ;r/6)  +  i  sm(- 5;r/6)]  •  2. 

7.  [3/5+i(-4/5)].5,  or  [cos(- 53°80  +  isin(- 53°80]  •  5. 

8.  (-  3/ V13  +  i '  2/ V13)  •  V13,  or  (cos  146°  19'  +  isin  146°  19')  •  V13. 

9.  [-3/5  +  i(-4/5)].10,  or[cos(-126°52')  +  isin(-126°520]-10. 

10.  [_V5/4  +  i(-Vll/4)]-4, 

or  [cos  (-  123°  590  +  i  sin  (-  123°  59')]  •  4. 

11.  cos  (90°  -<p)  +  i  sin  (90°  -  0). 

12.  cos  (90°  +  e)  +  i  sin  (90°  +  $). 

EXERCISE  XXX 

4.  Each  quality  unit  in  example  2  is  a  fifth  root  of  —  1.     Each  quality 
unit  in  example  3  is  a  sixth  root  of  —  1. 

5.  cos(6^  + 5  0) +  isin(6^4- 50). 

6.  cos(50/3- 3^/2) +  isin(50/3- 3^/2). 

7.  cos  (0  +  0  -  /3  -  7)  +  i  sin  (0  +  0  -  /3  -  7). 

8.  cos(— ;r)  +  zsin(— tt),  or  —  1. 

9.  cos(40  + 5^  -  7f/2)  +  isin(4  0  +  5^  -  ;f/2), 
or  sin  (4  0  +  5  ^)  -  i  cos  (40  +  5  6). 


168  PLANE  TRIGONOMETRY 

EXERCISE   XXXI 
1.    +1,  (-l±W3)/2.  2.    -1,  (l±V^)/2. 

3.  ±1,  (l±iV3)/2,  (-l±iV3)/2. 

4.  ±i,  (V3±i)/2,  (-V3±i)/2. 

5.  ±2,  ±i-2. 

6.  2,  2[cos(2n;r/5)  ±  isin(2n7r/5)],  where  n  =  1  or  2. 

Here  two  roots  are  the  reciprocals  of  two  others  respectively. 


7.  -3,  3[cos(2n  +  1  ;r/5)  + isin(2n  +  l;r/5)],  where  n  =  1,  2,3,  or  4. 


8.    d=  [cos(4n-  iTt/U)  +  isin(4n  -  1  tt/ 12)],  where  n  =  0,  1,  or  2. 
Here  the  roots  are  opposites  in  pairs. 


9.    2[cos(12n  +  l7r/18)  +  isin(12w  +  1 7r/18)],  wherew  =  0, 1,  or2. 


10.    ±[cos(6n-l7r/12)  +  isin(6n  -  l;r/12)]  V2,  where  w  =  0  or  1. 
Here  the  roots  are  opposites  in  pairs. 


11.  ±  [cos  (8  n  + 1  ;r/24)  +  i  sin  (8  w  + 1 7r/24)]  V2,  where  n  =  0,  1,  or  2. 
Here  the  roots  are  opposites  in  pairs. 

12.  v'4  [cos  (rTT / 1 5)  +  i  sin  (rTT/ 15)],  where  r  =  -  1,  5,  11,  17,  or  23. 

13.  cos(7r/5)  ±  *sin(7r/5),  cos(3;r/5)  ±  isin(3;r/5). 

14.  ±1,  (l±V^)/2,  (-1  ±iV3)/2,  ±i,  cos(;r/6)±isin(;r/6), 
cos(5;r/6)  ±  isin(5  7r/6). 

15.  -1,  (l±V33)/2,  ±[cos(^/4)  +  isin(;r/4)], 
±  [cos(3;r/4)  +  isin(3  7r/4)]. 

16.  -1,  cos(n^/7)  ±  isin(n;r/7),  where  w  =  1,  3,  or  5. 

EXERCISE   XXXn 

1.  n  •  360°  +  45°,  or  2  n;r  +  7r/4. 

2.  n.  360° +  132°.  4.    n •  360°  -  100°. 

3.  n  ■  360°  -  35°.  5.    2n7t  +  7t/6. 

6.   cos^  =  ±V33/7,  tan^  =  ±4/V33,  cot^  =  ±  V33/4, 
sec^  =  ±7/V33,  csc^  =  7/4. 


ANSWERS  169 

7.  cot  A  =  2/3,  sin^  =  ±3/V13,  cos^  =  ±2/V13, 
sec^  =  ±V13/2,  csc^  =  ±V13/3. 

8.  sin  ^  =  ±  V55/8,  tsuiA  =  T  V55/3,  cot^  =  =F  3/V5'5, 
sec  ^  =-  8/3,  CSC  J.  =±  S/^ob. 

9.  tanA  =  -  5/7,  sin^  =±  5/V74,  cos^  =T  7/V'J'4, 
sec  J.  =  T  V74/7,  csc^  =  ±  V74/5. 

10.  sm^  =  ±V33/7,  cos^  =  4/7,  tan^  =  ±V33/4, 
cot  ^  =  ±  4  /  V33,  esc  ^  =  ±  7  /  V33. 

11.  sin  ^=-4/5,  cosJ.  =  t3/5,  tan^  =±4/3,  cot^=±3/4, 
sec  ^  =T  5/3. 

12.  1st  or  2d  ;   1st  or  3d  ;  2d  or  3d  ;  2d  or  4th  ;  1st  or  4th  ;  3d  or  4th. 

13.  cos  4°.  16.    -  cot  40°  19.    -  cos  5°. 

14.  -  sin  38°.  17.    -  sec  10°.  20.    -  tan  20°. 

15.  tan  35°.  18.    -  cos  15°.  21.  tan  30°. 
22-27.    See  table  on  page  151.         66.   2d  or  3d. 

63.  2d  or  4th.  67.    UTt  ±7C/2. 

64.  1st  or  4th.  68.    n  ■  180°  +  ( -  1)"  21°  28'. 

65.  3d  or  4th.  69.    2n7t  ±  7t /2,  uTt  +  {- 1)»  tc/O. 

70.  n;r  +  (- l)»;r/6,  nTT -(- l)»;r/2. 

71.  n  •  180°  ±  35°  10'.       72.    2  njt  ±7t/3,  2  UTt  ±  it.      73.    nn  ±  ^t/i^. 

74.  n  •  360°  ±  51°  19',  n  •  360°  ±  180°. 

75.  nTT  +  (- l)«;r/6,  nTT  -  {- l)«;r/2. 

76.  mt  ±  7r/4. 

77.  n  ■  180°  +  45°,  n  •  180°  -  18°  26'. 

78.  n- 180°  +  (-  1)«30°,  n-  180°  -  (-  1)»19°28'. 

79.  n7r/2 +  (-  l)»7r/12,  ?i7r/2-(- l)«7r/4. 

80.  n7r  + (- l)«7r/6.  82.    2n7r  ±  ;r/4,  2  ttti  ±  7r/2. 

81.  n7r-(- l)"(7r/3).  83.    wtt  +  (- l)»;r/6. 


170  PLANE  TRIGONOMETRY 

84.  n  ■  180°,  n  •  360°  +  65°  43',  n  •  360°  +  204°  18^ 

85.  2n7r  ±7t/S,2n7t  ±7t/2.         86.    w  •  360°  +  105°,  n  •  360°  -  15° 
87.  2n7r  ±7t/2,2n7t,2n7t  ±27t/S. 


88.    2n7t  ±  the  principal  value  of  cos-^ ^ — 


89.  (2n+  l)7r/6. 

90.  n  ■  180°  +  22°  37',  n  •  180°  +  143°  8'. 

91.  ±1,  ±V21/7.  93.    1.  95.    V3/3,  -V3/2. 

92.  V3/2.  94.    V3/3.  96.    (n  +  l)7r/3. 

97.  rnt  ±  Tt / 6,  riTt  ±  7t / S. 

98.  n;r +  (-!)« 7r/6,  n  •  180°  +  (- 1)«  •  14°  29'. 

99.  2  uTt,  2  WTT  +  TT,  2  WTT  +  ;r/4,  2  utt  +  5  ;r/4. 
100.  riTT  ±  7t/S. 


101.  ^  =  tan-i  (a/b);  r  =  Va'^  +  62. 

102.  e  =  tan-i(a/6);  0  =  tan-i  (c/Va^  +  ft^);  r  =:  Va^  +  62  +  ^2). 


103.   a;  =  tan-i(a/6)  +  cos-i(±Va2_+62/2); 
2/ =  tan-i  (tt/6)  -  cos-i  (±  Va2  +  52/2). 


104.  X  =  tan-i  (-  b/a)  +  sin-i  (T  ^a^  +  &V2); 
2/  =  tan-'  (a/6)  -  sin-i(±  Va2  +  &V2). 

105.  X  =  (6sinB- acos5)/sin(^  -  J.); 
y  =  {a  cos  A  —  b  sin  ^)/sin  (5  —  ^). 


,  a  cos  5  -  6  sin  ^  Va2  _|_  52  _  2  a6  sm  {A  -  B) 

106.  ^  =  tan-i ^—  ;  r  = — -^ '  • 

6  cos  A  -\-  a  sni  B  cos  (^1  —  B) 

107.  x  =  0,  y  =  0;  x=7t,y  =  -  7t;  x  =  —  it,y  =  7t. 

108.  i2=Tr(sin/i  + cos/i);  F=  [IT- TF(sin/i  +  cos/i)cos/i]/sin/i. 


109.  X  =  r  vers-i  {y/r)T  v  2  ry  -  y^. 

110.  (ae  -  6d)2  =  (a/  -  cd)2  +  (ce  -  6/)2. 


ANSWERS  171 

111.   a-2  -f-  62  ^  c2  +  d2.  115.   4,  1  ±  V3. 

114.    -4,  2±2V3.  116.    -1,  (l±V3)/2. 

117.  -  1  +  2  cos  40°,  -  1  +  2  cos  160°,  -  1  +  2  cos  280°. 

118.  -  4/3  +  (2  V10/3)cos^,  where  A  =  39°  5'  51", 
159°  5'  51",  or  279°  5'  51". 

EXERCISE   XXXm 

1.  0.43498  mi.        3.  502.46  ft.         5.  71°  33'  54". 

2.  37°  36'  30".       4.  49°  44'  38".        6.  56.649  ft. 

7.  260.21  ft.,  3690.3  ft.  10.    238,850  mi. 

8.  235.81  yd.  11.    90,824,000  mi. 

20.  0.32149  mi.  24.    12.458  mi.  28.    278.7  ft. 

21.  942.7  yd.  25.    5.1083  mi.  29.    5422  yd. 

22.  5147.9  ft.  26.    1346.3  ft.  30.    28°  57'  20". 

23.  529.5  yd.  27.    0.83732  mi.  34.    210.4  ft. 

35.  9°  28' ;  2.9152  times  the  height  of  the  tower. 

36.  3121.1  ft.,  3633.5  ft.  38.    529.49  ft. 

39.  658.36  lb.,  22°  23'  47"  with  first  force. 

40.  88.326  lb.,  45°  37'  16"  with  known  force. 

41.  210  acres  9.1  sq.  ch.  42.    61  acres  4.97  sq.  ch. 

43.  1075.3.  47.    25.92  ft.  51.    3.7865. 

44.  16,281.  48.    33°  12' 4".  52.    1.9755. 

45.  749.95  sq.  ft.  49.    1.3764.  63.    6.4984. 

46.  1834.95  sq.  ft.  50.    36.024  ft.  54.    6.1981. 

55.    8.6058  sq.  ft. 


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